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:
# 閉路検出を繰り返し, 閉路の長さを減らしていく































def main():
    import sys
    input = sys.stdin.readline
    sys.setrecursionlimit(10**7)
    from collections import Counter, deque
    from itertools import combinations, permutations, accumulate, groupby, product
    from bisect import bisect_left,bisect_right
    from heapq import heapify, heappop, heappush
    import math
    #from math import gcd

    #inf = 10**17
    #mod = 10**9 + 7

    n,m = map(int, input().split())
    adj = [[] for _ in range(n)] #頂点数, 場合によって変える
    for _ in range(m):
        a,b = map(int, input().split())
        adj[a-1].append(b-1)


    def bfs(start):
        pre = [-1]*n
        d = deque([start])
        res = []

        while d:
            v = d.popleft()
            for nv in adj[v]:
                if nv == start:
                    path = []
                    while v != -1:
                        path.append(v+1)
                        v = pre[v]
                    if not res or len(res) > len(path):
                        res = path
                    break

                if pre[nv] != -1:
                    continue
                pre[nv] = v
                d.append(nv)

        return res
    
    res = [0]*(n+1)
    for i in range(n):
        temp = bfs(i)
        if temp and len(res) > len(temp):
            res = temp
    if len(res) <= n:
        print(len(res))
        print(*res, sep='\n')
    else:
        print(-1)

if __name__ == '__main__':
    main()