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;
const int N = 1e5 + 10;
vector<pair<int, int> > G[N];
vector<int> edge;
bool vis[N], fuck[N];
int instack[N];
pair<int, int> pre[N];
bool dfs(int u, bool flag = 0) {
  vis[u] = 1, instack[u] = 1;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i].first;
    int id = G[u][i].second;
    if (fuck[id]) continue;
    pre[v] = pair<int, int>(u, id);
    if (!vis[v]) {
      if (dfs(v, flag)) return 1;
    } else if (instack[v]) {
      if (!flag) return 1;
      edge.push_back(id);
      int cur = u;
      while (cur != v) {
        edge.push_back(pre[cur].second);
        cur = pre[cur].first;
      }
      return 1;
    }
  }
  instack[u] = 0;
  return 0;
}
void solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(pair<int, int>(v, i));
  }
  bool flag = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      if (dfs(i, 1)) {
        flag = 1;
        break;
      }
    }
  }
  if (!flag) {
    printf("YES\n");
    return;
  }
  for (int i = 0; i < edge.size(); i++) {
    fuck[edge[i]] = 1;
    bool flag = 1;
    memset(vis, 0, sizeof(vis));
    for (int j = 1; j <= n; j++) {
      if (!vis[j] && dfs(j)) {
        flag = 0;
        break;
      }
    }
    if (flag) {
      printf("YES\n");
      return;
    }
    fuck[edge[i]] = 0;
  }
  printf("NO\n");
}
int main() {
  solve();
  return 0;
}
