#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10, M = N << 1;
int ind[N];
int n, m;
//邻接表
int e[M], h[N], idx, ne[M];
void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
int q[N];
int main() {
    while (~scanf("%d %d", &n, &m) && n) {
        memset(ind, 0, sizeof(ind));
        memset(h, -1, sizeof h);
        idx = 0;

        for (int i = 1; i <= m; i++) {
            int a, b;
            scanf("%d %d", &a, &b);
            add(a, b);
            ind[b]++; //记录入度
        }
        int hh = 0, tt = -1;
        for (int i = 0; i < n; i++)
            if (ind[i] == 0) q[++tt] = i; //入度为0的入队列

        while (hh <= tt) {
            int u = q[hh++];
            for (int i = h[u]; ~i; i = ne[i]) {
                int j = e[i];
                ind[j]--;                     //拓扑模拟减去边，入度-1
                if (ind[j] == 0) q[++tt] = j; //如果入度为0，则此节点入队列
            }
        }
        //检查入过队列的节点个数是不是n个，相等则无环
        hh == n ? puts("YES") : puts("NO");
    }
    return 0;
}