Solve in Python:
You are given a 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 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 $2 \rightarrow 3$, and the graph becomes acyclic.

In the second example you have to remove at least two edges (for example, $2 \rightarrow 1$ and $2 \rightarrow 3$) 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):
    nonlocal color
    nonlocal 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():
    nonlocal color
    nonlocal 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')
    return

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

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

print('NO')