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 gcd1(int a, int b) {
  if (a == 0) return b;
  return gcd1(b % a, a);
}
long long modx(long long base, long long ex) {
  long long ans = 1LL, val = base;
  while (ex > 0LL) {
    if (ex & 1LL) ans = (ans * val) % 1000000009LL;
    val = (val * val) % 1000000009LL;
    ex = ex >> 1LL;
  }
  return ans;
}
const int maxn = 505;
set<int> adj[maxn], rev[maxn];
vector<int> v;
int parent[maxn];
int n, m, x, y;
bool recstack[maxn], visit[maxn];
int ind;
void cycle(int t) {
  while (t != parent[t]) {
    v.push_back(t);
    t = parent[t];
  }
  v.push_back(t);
}
bool dfs(int start, int par) {
  visit[start] = true;
  parent[start] = par;
  recstack[start] = true;
  set<int>::iterator it;
  for (it = adj[start].begin(); it != adj[start].end(); it++) {
    int pt = *it;
    ;
    if (recstack[pt]) {
      parent[pt] = pt;
      ind = start;
      return true;
    }
    if (!visit[pt] && dfs(pt, start)) return true;
  }
  recstack[start] = false;
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    adj[x].insert(y);
    rev[y].insert(x);
  }
  for (int i = 1; i <= n; i++) {
    if (!visit[i]) {
      if (!dfs(i, 0))
        continue;
      else {
        cycle(ind);
        break;
      }
    }
  }
  if (ind == 0) {
    cout << "YES" << endl;
    return 0;
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    int pt = v[i];
    int pt2;
    if (i == v.size() - 1)
      pt2 = v[0];
    else
      pt2 = v[i + 1];
    adj[pt].erase(pt2);
    int j = 1;
    memset(visit, 0, sizeof(visit));
    memset(recstack, 0, sizeof(recstack));
    for (j = 1; j <= n; j++) {
      if (!visit[j] && dfs(j, 0)) break;
    }
    if (j <= n) {
      adj[pt].insert(pt2);
      continue;
    } else {
      cout << "YES" << endl;
      return 0;
    }
  }
  cout << "NO" << endl;
  return 0;
}
