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;
template <typename T>
void read(T &x) {
  register T c = getchar();
  x = 0;
  int t = 0;
  if (c == '-') t = 1, c = getchar();
  for (; (c < 48 || c > 57); c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (t) x = -x;
}
vector<long long> g[505];
vector<long long> cycle;
long long till, from, to;
bool pushed[505], cycleFound, visited[505], mat[505][505];
void dfs1(int node) {
  if (cycleFound) return;
  cycle.push_back(node);
  visited[node] = 1;
  for (auto i : g[node]) {
    if (cycleFound) break;
    if (pushed[i]) {
      cycle.push_back(i);
      cycleFound = true;
      till = i;
      return;
    }
    if (visited[i]) continue;
    pushed[i] = true;
    dfs1(i);
  }
  if (cycleFound) return;
  cycle.pop_back();
  pushed[node] = 0;
}
void dfs2(int node) {
  visited[node] = 1;
  if (cycleFound) return;
  for (auto i : g[node]) {
    if (cycleFound) break;
    if (mat[node][i]) {
      if (pushed[i]) {
        cycleFound = true;
        return;
      }
      if (visited[i]) continue;
      pushed[i] = 1;
      dfs2(i);
    }
  }
  pushed[node] = 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, u, v;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    mat[u][v] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    if (!visited[i]) {
      pushed[i] = 1;
      dfs1(i);
    }
  }
  if (!cycleFound) {
    cout << "YES";
    exit(0);
  } else {
    long long i = 0;
    while (cycle[i] != till) {
      ++i;
    }
    for (; i + 1 < cycle.size(); ++i) {
      memset(visited, 0, sizeof(visited));
      memset(pushed, 0, sizeof(pushed));
      cycleFound = false;
      from = cycle[i];
      to = cycle[i + 1];
      mat[from][to] = 0;
      for (int j = 1; j <= n; ++j) {
        if (!visited[j]) {
          pushed[j] = 1;
          dfs2(j);
        }
      }
      mat[from][to] = 1;
      if (!cycleFound) {
        cout << "YES";
        return 0;
      }
    }
  }
  cout << "NO";
}
