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>
inline int read(T& x) {
  x = 0;
  char c = getchar(), f = 1;
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    if (c == -1) return -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  x *= f;
  return 1;
}
template <typename T, typename... Args>
inline int read(T& x, Args&... args) {
  return read(x) + read(args...);
}
template <typename T>
inline int write(T x) {
  if (x < 0) return putchar('-') & write((~x) + 1);
  if (x / 10) return write(x / 10) & putchar(x % 10 | 48);
  return putchar(x | 48);
}
template <typename T>
inline int write(T x, char c) {
  return write(x) + putchar(c);
}
int n, m;
struct edge {
  int u, v, nxt;
} e[100005];
int f[505], tot;
inline void addEdge(int u, int v) {
  tot++, e[tot].u = u, e[tot].v = v, e[tot].nxt = f[u], f[u] = tot;
}
int indeg[505];
int tdeg[505];
inline bool check() {
  int idx = 0;
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (!indeg[i]) q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    idx++;
    for (int i = f[x]; i; i = e[i].nxt) {
      indeg[e[i].v]--;
      if (!indeg[e[i].v]) q.push(e[i].v);
    }
  }
  return idx == n;
}
signed main() {
  read(n, m);
  for (int i = 1, u, v; i <= m; i++) {
    read(u, v);
    addEdge(u, v);
    indeg[v]++;
    tdeg[v] = indeg[v];
  }
  for (int i = 1; i <= n; i++) {
    if (indeg[i]) {
      indeg[i]--;
      if (check()) return puts("YES") & 0;
      for (int j = 1; j <= n; j++) indeg[j] = tdeg[j];
    }
  }
  puts("NO");
}
