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 = 1e5 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
struct graph {
  int head[maxn], nxt[maxn << 1], to[maxn << 1], w[maxn << 1], sz;
  void init() { memset(head, -1, sizeof(head)); }
  graph() { init(); }
  void push(int a, int b, int c = 0) {
    nxt[sz] = head[a], to[sz] = b, w[sz] = c, head[a] = sz++;
  }
  int& operator[](const int a) { return to[a]; }
} g;
set<int> s[505];
int deg[505];
int tdeg[505];
bool vis[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1, a, b; i <= m; ++i) {
    scanf("%d%d", &a, &b);
    g.push(a, b);
    deg[b]++;
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    if (!deg[i]) q.push(i);
  }
  int cnt = 0;
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    vis[now] = 1;
    cnt++;
    for (int i = g.head[now]; ~i; i = g.nxt[i]) {
      deg[g[i]]--;
      if (deg[g[i]] == 0) q.push(g[i]);
    }
  }
  if (cnt == n) {
    printf("YES\n");
    return 0;
  }
  bool ok = false;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      vis[i] = 1;
      q.push(i);
      int c2 = 0;
      for (int i = 1; i <= n; ++i) tdeg[i] = deg[i];
      while (!q.empty()) {
        int now = q.front();
        q.pop();
        c2++;
        for (int i = g.head[now]; ~i; i = g.nxt[i]) {
          if (vis[g[i]]) continue;
          tdeg[g[i]]--;
          if (tdeg[g[i]] == 0) q.push(g[i]);
        }
      }
      if (c2 + cnt == n) {
        ok = true;
        break;
      }
      vis[i] = 0;
    }
  }
  if (ok)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
