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 maxn = 500;
vector<int> G[maxn + 5], G2[maxn + 5];
vector<int> s;
bool vis[maxn + 5];
vector<int> G3[maxn + 5], G4[maxn + 5];
queue<int> q;
int cc[maxn + 5];
int sz, cp;
int cur_cc;
int n, m, x, y;
int cnt[maxn + 5], in[maxn + 5], deg[maxn + 5];
void dfs1(int u) {
  vis[u] = true;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (!vis[v]) dfs1(v);
  }
  s.push_back(u);
}
void dfs2(int u) {
  cc[u] = cur_cc;
  cnt[cur_cc]++;
  for (int i = 0; i < G2[u].size(); i++) {
    int v = G2[u][i];
    if (!cc[v]) dfs2(v);
  }
}
void find_scc() {
  cur_cc = 0;
  s.clear();
  memset(vis, false, sizeof(vis));
  memset(cc, 0, sizeof(cc));
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs1(i);
  for (int i = n - 1; i >= 0; i--)
    if (!cc[s[i]]) {
      cur_cc++;
      dfs2(s[i]);
    }
}
bool check() {
  memset(in, 0, sizeof(in));
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < G4[i].size(); j++) {
      int v = G4[i][j];
      in[v]++;
    }
  while (q.size()) q.pop();
  int tem = 0;
  for (int i = 1; i <= n; i++)
    if (in[i] == 0 && cc[i] == cp) q.push(i);
  while (q.size()) {
    int u = q.front();
    q.pop();
    tem++;
    for (int i = 0; i < G4[u].size(); i++) {
      int v = G4[u][i];
      in[v]--;
      if (in[v] == 0) q.push(v);
    }
  }
  return tem == sz;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G2[y].push_back(x);
  }
  find_scc();
  int num = 0;
  for (int i = 1; i <= cur_cc; i++)
    if (cnt[i] > 1) num++;
  if (num >= 2)
    printf("NO\n");
  else if (num == 0)
    printf("YES\n");
  else {
    cp = -1;
    for (int i = 1; i <= cur_cc; i++)
      if (cnt[i] > 1) {
        cp = i;
        break;
      }
    sz = cnt[cp];
    for (int i = 1; i <= n; i++) {
      if (cc[i] != cp) continue;
      for (int j = 0; j < G[i].size(); j++) {
        int v = G[i][j];
        if (cc[v] == cp) G3[i].push_back(v);
      }
    }
    memset(deg, 0, sizeof(deg));
    for (int i = 1; i <= n; i++)
      for (int j = 0; j < G3[i].size(); j++) deg[G3[i][j]]++;
    for (int i = 1; i <= n; i++)
      for (int j = 0; j < G3[i].size(); j++) {
        int v = G3[i][j];
        if (deg[v] == 1) {
          for (int i1 = 1; i1 <= n; i1++) G4[i1].clear();
          for (int i1 = 1; i1 <= n; i1++)
            for (int j1 = 0; j1 < G3[i1].size(); j1++)
              if (i1 == i && G3[i1][j1] == v)
                continue;
              else
                G4[i1].push_back(G3[i1][j1]);
          if (check()) {
            printf("YES\n");
            return 0;
          }
        }
      }
    printf("NO\n");
  }
  return 0;
}
