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:
from collections import defaultdict
import heapq


def main():
    N, M = list(map(int, input().split()))
    adj = defaultdict(list)
    for _ in range(M):
        A, B = list(map(lambda x: int(x) - 1, input().split()))
        adj[A].append(B)
    # 最小サイズのサイクルを探す
    min_cycle_size = N + 1
    min_cycle = list()
    for s in range(N):
        parent = [None] * N
        que = [(0, s)]
        while len(que) > 0:
            dist, node = heapq.heappop(que)
            if dist > 0 and node == s:
                break
            for next_node in adj[node]:
                if parent[next_node] is not None:
                    continue
                parent[next_node] = node
                heapq.heappush(que, (dist + 1, next_node))
        # s始まりの最小サイクルが見つかった場合
        if parent[s] is not None:
            cycle = [s]
            node = parent[s]
            while node != s:
                cycle.append(node)
                node = parent[node]
            if len(cycle) < min_cycle_size:
                min_cycle_size = len(cycle)
                min_cycle = cycle
    # 答え
    if min_cycle_size > N:
        print(-1)
    else:
        print(min_cycle_size)
        for node in min_cycle:
            print(node + 1)


if __name__ == '__main__':
    main()