#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 100010, M = 300010;
//与AcWing 1169. 糖果 这道题一模一样，连测试用例都一样

stack<int> q; //有时候换成栈判断环很快就能
LL dist[N];
bool st[N];
int cnt[N];
int n, m; //表示点数和边数
//邻接表
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

bool spfa() {                         //求最长路，所以判断正环
    memset(dist, -0x3f, sizeof dist); //初始化为-0x3f
    //差分约束从超级源点出发
    dist[0] = 0;
    q.push(0);
    st[0] = true;

    while (q.size()) {
        int u = q.top();
        q.pop();
        st[u] = false;
        for (int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            if (dist[j] < dist[u] + w[i]) { //求最长路
                dist[j] = dist[u] + w[i];
                cnt[j] = cnt[u] + 1;
                //注意多加了超级源点到各各节点的边
                if (cnt[j] >= n + 1) return false;
                if (!st[j]) {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    return true;
}

int main() {
    memset(h, -1, sizeof h);
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int op, a, b; // op为选择
        scanf("%d %d %d", &op, &a, &b);
        if (op == 1) /** a == b  =>  (a >= b , b >= a)  */
            add(a, b, 0), add(b, a, 0);
        else if (op == 2) /**  b >= a + 1         */
            add(a, b, 1);
        else if (op == 3) /**  a >= b         */
            add(b, a, 0);
        else if (op == 4) /**  a >= b + 1        */
            add(b, a, 1);
        else /** b >= a   */
            add(a, b, 0);
    }

    /** xi >= x0 + 1 （每个小朋友都要至少一个糖果）*/
    //将所有节点与超级源点x0相连
    for (int i = 1; i <= n; ++i) add(0, i, 1);

    if (!spfa())
        puts("-1");
    else {
        LL res = 0;
        for (int i = 1; i <= n; ++i) res += dist[i];
        printf("%lld\n", res);
    }
    return 0;
}