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.
cycle_begin, cycle_end = -1, -1
g, mark, prev, edges = [], [], [], []

def dfs(u):
    global cycle_begin, cycle_end
    mark[u] = 1
    for v in g[u]:
        if mark[v] == 0:
            prev[v] = u
            if dfs(v):
                return True
        elif mark[v] == 1:
            cycle_begin = v
            cycle_end = u
            return True
    mark[u] = 2
    return False

def dfs2(u):
    mark[u] = 1
    for v in g[u]:
        if v != -1:
            if mark[v] == 0:
                if dfs2(v):
                    return True
            elif mark[v] == 1:
                return True
    mark[u] = 2
    return False

n, m = map(int, input().split())
g = [[] for i in range(n)]
mark = [0 for i in range(n)]
prev = [-1 for i in range(n)]
for i in range(m):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    g[u].append(v)
for i in range(n):
    if mark[i] == 0 and dfs(i):
        break
if cycle_begin == -1:
    print("YES")
else:
    cycle = []
    i = cycle_end
    while i != cycle_begin:
        cycle.append(i)
        i = prev[i]
    cycle.append(cycle_begin)
    cycle.reverse()
    edges = []
    for i in range(len(cycle) - 1):
        edges.append(tuple((cycle[i], cycle[i + 1])))
    edges.append(tuple((cycle[len(cycle) - 1], cycle[0])))
    can = False
    while len(edges) > 0:
        f = edges[0][0]
        s = edges[0][1]
        g[f][g[f].index(s)] = -1
        mark = [0 for i in range(n)]
        have = False
        for i in range(n):
            if mark[i] == 0 and dfs2(i):
                have = True
                break
        g[f][g[f].index(-1)] = s
        if not have:
            can = True
            break
        edges.pop(0)
    if can:
        print("YES")   
    else:
        print("NO")
