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():
    n,m=map(int,input().split())
    ab=[list(map(int,input().split())) for _ in [0]*m]
    g=[set() for _ in [0]*n]
    g2=[set() for _ in [0]*n]
    [g[a-1].add(b-1) for a,b in ab]
    [g2[b-1].add(a-1) for a,b in ab]

    def bfs(start,graph):
        temp=[[start,{start}]]
        while temp:
            temp2=[]
            for j,k in temp:
                for i in graph[j]:
                    if i in k:
                        return j,k
                    else:
                        temp2.append([i,k|{i}])
            temp=temp2
        return None,None

    return_set_pre=set(range(n))
    for i in range(n):
        now,return_set=bfs(i,g)
        if return_set!=None:
            break
    if return_set==None:
        print(-1)
        return 0
    while True:
        for i in return_set_pre-return_set:
            g[i]={}
            g2[i]={}
        for i in return_set:
            g[i]=g[i]&return_set
            g2[i]=g2[i]&return_set
        l1=len([1 for i in range(n) if len(g[i])==1])
        l2=len([1 for i in range(n) if len(g2[i])==1])
        l=len(return_set)
        if l==l1 and l==l2:
            break
        return_set_pre=return_set
        now,return_set=bfs(now,g2)
        for i in return_set_pre-return_set:
            g[i]={}
            g2[i]={}
        for i in return_set:
            g[i]=g[i]&return_set
            g2[i]=g2[i]&return_set
        l1=len([1 for i in range(n) if len(g[i])==1])
        l2=len([1 for i in range(n) if len(g2[i])==1])
        l=len(return_set)
        if l==l1 and l==l2:
            break
        return_set_pre=return_set
        now,return_set=bfs(now,g)
    print(len(return_set))
    for i in return_set:
        print(i+1)
main()