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 deque
import sys
input=sys.stdin.readline


def solve():
    N, M = map(int, input().split())
    d = {i:[] for i in range(1, N+1)}
    l = []
    INF = 10**12
    v = INF

    for _ in range(M):
        A, B = map(int, input().split())
        d[A].append(B)
        l.append((A, B))

    ans = []
    for i, j in l:
        INF = 10**12
        dist = [-1]*(N+1)
        prev = [-1]*(N+1)
        dist[j] = 0
        prev[j] = 0
        q = deque([j])

        while q:
            a = q.popleft()
            for b in d[a]:
                if dist[b]==-1:
                    dist[b] = dist[a]+1
                    prev[b] = a
                    q.append(b)

        if v>dist[i]+1 and dist[i]!=-1:
            v = dist[i]+1
            buf = i
            ans = []
            while buf:
                ans.append(buf)
                buf = prev[buf]

    if ans:
        print(len(ans))
        for i in ans:
            print(i)
    else:
        print(-1)


if __name__ == "__main__":
    solve()