Problem: Given is a directed graph G with N vertices and M edges.
The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i.
It is guaranteed that the graph contains no self-loops or multiple edges.

Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph.
Here the null graph is not considered as a subgraph.

Constraints

* 1 \leq N \leq 1000
* 0 \leq M \leq 2000
* 1 \leq A_i,B_i \leq N
* A_i \neq B_i
* All pairs (A_i, B_i) are distinct.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
A_2 B_2
:
A_M B_M


Output

If there is no induced subgraph of G that satisfies the condition, print `-1`. Otherwise, print an induced subgraph of G that satisfies the condition, in the following format:


K
v_1
v_2
:
v_K


This represents the induced subgraph of G with K vertices whose vertex set is \\{v_1, v_2, \ldots, v_K\\}. (The order of v_1, v_2, \ldots, v_K does not matter.) If there are multiple subgraphs of G that satisfy the condition, printing any of them is accepted.

Examples

Input

4 5
1 2
2 3
2 4
4 1
4 3


Output

3
1
2
4


Input

4 5
1 2
2 3
2 4
1 4
4 3


Output

-1


Input

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2


Output

4
2
3
4
5

Solution (in PYTHON3):

Solution:
#!/usr/bin/env python

import sys
sys.setrecursionlimit(10 ** 6)
from collections import deque
input = sys.stdin.buffer.readline
INF = float("inf")

def main():
    N,M = map(int,input().split())
    G = [[] for _ in range(N)]
    AB = [tuple(map(int,input().split())) for _ in range(M)]
    for a,b in AB:
        G[a-1].append(b-1)

    Ans = [1]*(N+1)
    
    # 全ての頂点から, 閉路が存在しないかを確認する
    for s in range(N):
        prev = [-1]*N
        dist = [-1]*N
        dist[s] = 0
        dq = deque([s])
        while dq:
            v = dq.popleft()
            for nv in G[v]:
                if dist[nv] == -1:
                    dist[nv] = dist[v] + 1
                    prev[nv] = v
                    dq.append(nv)
                # 次が 0番目(開始地点) なら閉路あり
                if dist[nv] == 0 and prev[nv] == -1:
                    prev[nv] = v

        # 開始地点の前 が -1でない => 閉路あり
        if prev[s] != -1:
            v = prev[s]
            ans = set()
            for _ in range(max(dist)+1):
                ans.add(prev[v])
                v = prev[v]
            if len(Ans) > len(ans):
                Ans = list(ans)
    
    if len(Ans) == N+1:
        print(-1)
    else:
        print(len(Ans))
        Ans.sort()
        for v in Ans:
            print(v+1)

if __name__ == "__main__":
    main()