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#include <bits/stdc++.h>
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using namespace std;
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const int N = 2e5 + 10, M = 2e6 + 10;
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//链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int dfn[N], low[N], stk[N], top, timestamp;
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vector<int> bcc[N];
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int bcnt;
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unordered_set<int> in;
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int n, m;
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int ans1, ans2, ans3;
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void deal_circle() {
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int sum = 0;
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for (int u : bcc[bcnt])
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for (int j = h[u]; ~j; j = ne[j])
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if (in.count(e[j])) sum++; //此边的两个点都在点双中，则此边在点双中
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sum /= 2; //因为a->b,b->a都sum++,重复了，需要除以2,才是边数
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//边数大于点数，嵌套环，每条边都是冲突边
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if (sum > bcc[bcnt].size()) ans2 += sum;
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//判定割边 (第二种判定割边的办法)
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//此处写ans3+=sum 或 ans3+=1是一个意思，因为此时只能是两点一线的情况，否则就不是点双。特殊的点双只有一条割边
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if (sum < bcc[bcnt].size()) ans3 += sum;
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}
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void tarjan(int u, int fa) {
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low[u] = dfn[u] = ++timestamp;
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stk[++top] = u;
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int son = 0; //子节点个数
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (j == fa) continue;
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if (!dfn[j]) {
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son++; //找到一个子节点
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tarjan(j, u);
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low[u] = min(low[u], low[j]);
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//第一种判定割边的办法
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//参阅一下判割边的模板代码
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if (low[j] > dfn[u]) ans1++;
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if (low[j] >= dfn[u]) {
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int x;
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bcnt++;
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//针对每个点双而言，都在dfs过程中记录此点双中有哪些点，避免后期二次统计，造成速度下降TLE
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in.clear();
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do {
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x = stk[top--];
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bcc[bcnt].push_back(x);
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in.insert(x); //增加点双中有哪些点的记录信息
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} while (x != j);
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bcc[bcnt].push_back(u);
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in.insert(u); //增加点双中有哪些点的记录信息
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deal_circle(); //在完成点双的记录后，马上开始统计，以免将所有的点双中点的信息完整记录下来，造成MLE,TLE等问题
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}
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} else
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low[u] = min(low[u], dfn[j]);
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}
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//特判独立点
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if (fa == -1 && son == 0) bcc[++bcnt].push_back(u);
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}
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int main() {
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//文件输入输出
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#ifndef ONLINE_JUDGE
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freopen("HDU3394.in", "r", stdin);
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#endif
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while (~scanf("%d%d", &n, &m) && n + m) {
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// ans1:割边数量 ans2:冲突边数量，也就是点双中所有边的数量，它们都是存在多个环中 ans3:也是割边的数量，与 ans1相同，但实现的方式不一样
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ans1 = ans2 = ans3 = idx = top = timestamp = 0;
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memset(dfn, 0, sizeof dfn);
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memset(h, -1, sizeof h);
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// memset(low, 0, sizeof low); //low数组不用重新初始化
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while (m--) {
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int a, b;
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scanf("%d %d", &a, &b);
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a++, b++; //避开下标0，向右推一个，使得下面的 tarjan算法的模板不变
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if (a != b) add(a, b), add(b, a); //防止重边
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}
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// tarjan求点双的模板
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i, -1);
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//下面无论是取 ans1还是ans3都是一样的
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printf("%d %d\n", ans1, ans2);
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// printf("%d %d\n", ans3, ans2);
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}
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return 0;
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}
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