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
from copy import deepcopy as copy
from collections import deque
input = sys.stdin.readline

def main():
    n, m = map(int, input().split())
    
    tree = [[] for _ in range(n)]
    edge = dict()
    for i in range(m):
        a, b = map(int, input().split())
        tree[a-1].append(b-1)
        edge[(a-1, b-1)] = i+1
    
    ans = 0
    for i in range(n):
        not_yet = deque([i])
        already = [False]*n
        dist = [0]*n
        par = [0]*n
        already[i] = True
        sub = []
        
        while not_yet:
            key = not_yet.popleft()
            for v in tree[key]:
                if already[v]:
                    
                    if v == i:
                        sub = [key+1]
                        now = key
                        while now != i:
                            sub.append(par[now]+1)
                            now = par[now]
                        break
                    continue
                
                not_yet.append(v)
                already[v] = True
                dist[v] = dist[key] + 1
                par[v] = key
            if sub:
                break
            
        if sub:
            if ans:
                if len(ans) > len(sub):
                    ans = copy(sub)
            else:
                ans = copy(sub)
            
        
    if ans:
        print(len(ans))
        for a in ans:
            print(a)
    else:
        print(-1)
        
    
    
if __name__ == "__main__":
    main()