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;
struct edge {
  int u, v;
  edge() {}
  edge(int u, int v) : u(u), v(v) {}
};
vector<edge> cycle;
vector<int> vec[505];
int col[505], par[505], mark;
void dfs1(int s) {
  int i, j, k, p;
  col[s] = 1;
  for (i = 0; i < vec[s].size(); i++) {
    if (cycle.size()) return;
    j = vec[s][i];
    if (col[j] == 1) {
      k = s;
      cycle.push_back(edge(k, j));
      while (par[k] != j) {
        cycle.push_back(edge(par[k], k));
        k = par[k];
      }
      cycle.push_back(edge(j, k));
      return;
    } else if (col[j] == 0) {
      par[j] = s;
      dfs1(j);
    }
  }
  col[s] = 2;
}
void dfs2(int s, int u, int v) {
  int i, j, k, p;
  col[s] = 1;
  for (i = 0; i < vec[s].size(); i++) {
    if (mark) return;
    j = vec[s][i];
    if (s == u && j == v) continue;
    if (col[j] == 1) {
      mark = 1;
      return;
    } else if (col[j] == 0)
      dfs2(j, u, v);
  }
  col[s] = 2;
}
void prexit() {
  printf("YES\n");
  exit(0);
}
int main() {
  int n, m, i, j, k, p, q, x, y;
  scanf("%d %d", &n, &m);
  for (i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    vec[x].push_back(y);
  }
  for (i = 1; i <= n; i++) {
    if (col[i] == 0) {
      dfs1(i);
      if (cycle.size()) break;
    }
  }
  if (cycle.size() == 0) prexit();
  for (j = 0; j < cycle.size(); j++) {
    memset(col, 0, sizeof(col));
    mark = 0;
    int u = cycle[j].u, v = cycle[j].v;
    for (i = 1; i <= n; i++) {
      if (col[i] == 0) {
        dfs2(i, u, v);
        if (mark) break;
      }
    }
    if (mark == 0) prexit();
  }
  printf("NO\n");
  return 0;
}
