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:
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)

    # 始点, 今回見る点
    def dfs(s, v):
        for nv in adj[v]:
            if nv == s:
                return [v]
            if visited[nv] == 1:
                continue
            visited[nv] = 1
            res = dfs(s, nv)
            if res != -1:
                res.append(v)
                return res
        return -1


    for s in range(n):
        visited = [0]*n
        visited[s] = 1
        heiro = dfs(s, s)
        if heiro != -1:
            break
    else:
        print(-1)
        exit()
    heiro = heiro[::-1]

    def graph(s, g):
        res = [s]
        d = deque([s])
        for i in range(len(heiro)):
            if heiro[i] == s:
                idx = i
                break
        while d:
            v = d.popleft()
            for nv in adj[v]:
                if visited[nv] == 1 and nv == g:
                    res.append(nv)
                    return res
            d.append(heiro[idx+1])
            res.append(heiro[idx+1])
            idx += 1
            if idx == len(heiro)-2:
                idx = -1
            #ここ
            if res[-1] == heiro[heiro.index(s)-1]:
                return res

    while True:
        heiro.append(heiro[0])
        visited = [0]*n
        f = 0
        for i in heiro:
            visited[i] = 1
        new_heiro = [heiro[0]]
        for i in range(len(heiro)-1):
            for nv in adj[heiro[i]]:
                if visited[nv] == 1 and nv != heiro[i+1]:
                    new_heiro = graph(nv, heiro[i])
                    f = 1
                    break
            else:
                new_heiro.append(heiro[len(new_heiro)])
            if f:
                heiro = new_heiro
                break
        else:
            break

    print(len(heiro)-1)
    for i in heiro[:-1]:
        print(i+1)

if __name__ == '__main__':
    main()