You are given a [directed graph](https://en.wikipedia.org/wiki/Directed_graph) consisting of n vertices and m edges (each edge is directed, so it can be traversed in only one direction). You are allowed to remove at most one edge from it.

Can you make this graph [acyclic](https://en.wikipedia.org/wiki/Directed_acyclic_graph) by removing at most one edge from it? A directed graph is called acyclic iff it doesn't contain any cycle (a non-empty path that starts and ends in the same vertex).

Input

The first line contains two integers n and m (2 ≤ n ≤ 500, 1 ≤ m ≤ min(n(n - 1), 100000)) — the number of vertices and the number of edges, respectively.

Then m lines follow. Each line contains two integers u and v denoting a directed edge going from vertex u to vertex v (1 ≤ u, v ≤ n, u ≠ v). Each ordered pair (u, v) is listed at most once (there is at most one directed edge from u to v).

Output

If it is possible to make this graph acyclic by removing at most one edge, print YES. Otherwise, print NO.

Examples

Input

3 4
1 2
2 3
3 2
3 1


Output

YES


Input

5 6
1 2
2 3
3 2
3 1
2 1
4 5


Output

NO

Note

In the first example you can remove edge <image>, and the graph becomes acyclic.

In the second example you have to remove at least two edges (for example, <image> and <image>) in order to make the graph acyclic.
#include <bits/stdc++.h>
using namespace std;
int n, m;
vector<int> g[505];
int ind[505];
queue<int> q;
struct node {
  int u, v;
};
bool topo(int v, int inds[], int cnt) {
  int tmp[505];
  for (int i = 0; i < 505; i++) tmp[i] = inds[i];
  tmp[v]--;
  if (tmp[v]) return false;
  while (!q.empty()) q.pop();
  q.push(v);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    cnt++;
    for (int i = 0; i < g[x].size(); i++) {
      tmp[g[x][i]]--;
      if (!tmp[g[x][i]]) q.push(g[x][i]);
    }
  }
  if (cnt == n) return true;
  return false;
}
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    while (!q.empty()) q.pop();
    memset(ind, 0, sizeof(ind));
    for (int i = 0; i < 505; i++) g[i].clear();
    int u, v;
    for (int i = 0; i < m; i++) {
      u = read();
      v = read();
      g[u].push_back(v);
      ind[v]++;
    }
    for (int i = 1; i <= n; i++)
      if (!ind[i]) q.push(i);
    int cnt = 0;
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      cnt++;
      for (int i = 0; i < g[x].size(); i++) {
        ind[g[x][i]]--;
        if (!ind[g[x][i]]) q.push(g[x][i]);
      }
    }
    if (cnt == n) {
      printf("YES\n");
      continue;
    }
    bool flag = false;
    for (int i = 1; i <= n; i++) {
      if (ind[i]) {
        if (topo(i, ind, cnt)) {
          flag = true;
          printf("YES\n");
          break;
        }
      }
    }
    if (!flag) printf("NO\n");
  }
}
