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

color = [0] * n # 0 - white, 1 - grey, 2 - black
cycle = []
blocked_u, blocked_v = -1, -1

def dfs(u):
    global color
    global cycle
    if color[u]:
        return
    color[u] = 1
    for v in a[u]:
        if u == blocked_u and v == blocked_v:
            continue
        if color[v] == 0:
            dfs(v)
        if color[v] == 1 or cycle:
            if not(cycle):
                cycle.append(v)
            cycle.append(u)
            return True
    color[u] = 2
    return False

def find_cycle():
    global color
    global cycle
    color = [0] * n # 0 - white, 1 - grey, 2 - black
    cycle = []
    for u in range(n):
        if dfs(u):
            break
    result = cycle[::-1]
    return {(result[i], result[(i + 1) % len(result)]) for i in range(len(result))}

cur = find_cycle()
if not(cur):
    print('YES')
    exit()

for bu, bv in cur:
    blocked_u = bu
    blocked_v = bv
    new = find_cycle()

    if not(new):
        print('YES')
        exit()

print('NO')
