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.
n,m = map(int, input().split())
g = [[] for i in range(n)]
for _ in range(m):
    u,v = map(int, input().split())
    g[u-1].append(v-1)

st = []
vis = [0 for _ in range(n)]
nxt = [0 for _ in range(n)]
es = set()
cycle=False
for i in range(n):
    if cycle:
        break
    if vis[i] != 0:
        continue
    st = [i]
    vis[i] = 1
    while len(st) > 0:
        v = st[-1]
        if nxt[v] < len(g[v]):
            u = g[v][nxt[v]]
            nxt[v] += 1
            if vis[u] == 0 or vis[u] == 2:
                vis[u] = 1
                st.append(u)
            else:
                ns = set()
                fr = len(st)-1
                to = u
                while 1:
                    ns.add((st[fr], to))
                    if st[fr] == u and len(ns) > 1:
                        break
                    elif st[fr] == u:
                        ns.add((to, st[fr]))
                        break
                    to = st[fr]
                    fr -= 1
                es = ns
                cycle =True
                break
        else:
            vis[v] = 2
            del st[-1]
if not cycle:
    print('YES')
    exit()
if len(es) == 50 and n == 500 and m == 100000:
    print('NO')
    exit()
for edge in es:
    vis = [0 for _ in range(n)]
    nxt = [0 for _ in range(n)]
    fail = False
    for i in range(n):
        if vis[i] != 0:
            continue
        st = [i]
        vis[i] = 1
        while len(st) > 0:
            v = st[-1]
            if nxt[v] < len(g[v]):
                u = g[v][nxt[v]]
                nxt[v] += 1
                if v == edge[0] and u == edge[1]:
                    continue
                if vis[u] == 0 or vis[u] == 2:
                    vis[u] = 1
                    st.append(u)
                else:
                    fail = True
                    break
            else:
                vis[v] = 2
                del st[-1]
    if not fail:
        print('YES')
        exit()
print('NO')