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#include <bits/stdc++.h>
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// https://www.cnblogs.com/yym2013/p/3845448.html
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// https://www.cnblogs.com/zhxmdefj/p/11117791.html#4000267478
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// 董晓老师讲解 按秩合并并查集
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// https://www.bilibili.com/video/av838604930
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/**
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按秩合并的并查集
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Q:为什么要将数量小的向数量大的连边，而不是倒过来？
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A:1.通过比较并查集的大小来连边，将小的(u)向大的(v)连边，这样对于大的并查集，查询代价不变，而小的并查集查询代价每个点增加了1，
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相当于增加了siz[u],(反过来则变成了siz[v]),siz[u]<siz[v]所以这样更优。
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2.相较于路径压缩的并查集，他可以保留原始的信息。
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*/
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/**
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// 上面是一采用递归的方式压缩路径， 但是，递归压缩路径可能会造成溢出栈，我曾经因为这个RE了n次，
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下面我们说一下非递归方式进行的路径压缩：
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int find(int x){
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int k, j, r;
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r = x;
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while(r != parent[r]) //查找跟节点
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r = parent[r]; //找到跟节点，用r记录下
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k = x;
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while(parent[k] != r) //非递归路径压缩操作
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{
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j = parent[k]; //用j暂存parent[k]的父节点
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parent[k] = r; //parent[x]指向跟节点
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k = j; //k移到父节点
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}
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return r; //返回根节点的值
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}
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*/
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// https://www.cnblogs.com/ARTlover/p/9752355.html
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using namespace std;
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const int N = 30001;
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int s[N]; //每个集合人数
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int fa[N]; //并查集数组
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int rnk[N]; //树高,不能使用rank变量名，与std中的已定义概念冲突
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/**
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测试数据
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100 4
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2 1 2
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5 10 13 11 12 14
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2 0 1
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2 99 2
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200 2
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1 5
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5 1 2 3 4 5
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1 0
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0 0
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答案：
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4
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1
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1
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*/
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//查询+带路径压缩
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int find(int x) {
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if (x == fa[x]) return x;
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else return fa[x] = find(fa[x]); //带路径压缩
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}
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//合并并查集
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void join(int x, int y) {
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//找出双方的根
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int xRoot = find(x);
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int yRoot = find(y);
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//同根则结束
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if (xRoot == yRoot) return;
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//让rank比较高的作为父结点
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if (rnk[xRoot] > rnk[yRoot]) {
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fa[yRoot] = xRoot;
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s[xRoot] += s[yRoot];//人数也需要合并进去
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} else {//小于等于都进这里~
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fa[xRoot] = yRoot;
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if (rnk[xRoot] == rnk[yRoot]) rnk[yRoot]++; //树的高度是在这里修改的，是什么意思？看董晓老师的视频，讲解的很清楚
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s[yRoot] += s[xRoot];//人数也需要合并进去
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}
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}
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int n, m, k, x, y;
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int main() {
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while (cin >> n >> m) {
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//输入为 0 0 结束
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if (n == 0 && m == 0) return 0;
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//初始化
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for (int i = 0; i < n; i++) { //因为有0号学生，所以数组遍历从0开始
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fa[i] = i; //每个学生自己是自己的祖先
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s[i] = 1; //集合中总人数，初始值是1
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rnk[i] = 0; //树的高度为0
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}
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for (int i = 0; i < m; i++) {
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cin >> k >> x;
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k--;
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while (k--) {
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cin >> y;
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join(x, y);
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}
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}
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cout << s[find(0)] << endl; //寻找0号学生相关集合的总人数
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}
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return 0;
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}
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