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:
import sys
sys.setrecursionlimit(10**7)

def main():
    N, M = map(int, input().split())
    V = [[] for _ in range(N)]
    for _ in range(M):
        a, b = map(int, input().split())
        V[a - 1].append(b - 1)
    C = [0] * N
    loop = []
    def dfs(n):
        if C[n] == 2:
            return False
        if C[n] == 1:
            loop.append(n)
            return True
        C[n] = 1
        for i in V[n]:
            if dfs(i):
                loop.append(n)
                return True
        C[n] = 2
        return False
    for i in range(N):
        if dfs(i):
            break
    if len(loop) == 0:
        print('-1')
        return
    for i in range(1, len(loop)):
        if loop[i] == loop[0]:
            break
    loop = loop[:i][::-1]

    def minimize(l):
        s = set(l)
        for i in range(len(l)):
            for j in V[l[i]]:
                if j == l[(i + 1) % len(l)]:
                    continue
                if j in s:
                    break
            else:
                continue
            break
        else:
            return l
        I = {m: k for k, m in enumerate(l)}
        j = I[j]
        if i < j:
            l = l[0:i + 1] + l[j:]
        else:
            l = l[j:i + 1]
        return l
    while True:
        t = len(loop)
        loop = minimize(loop)
        if len(loop) == t:
            break
    print(t)
    print('\n'.join(str(i + 1) for i in loop))
main()