p02902 AtCoder Beginner Contest 142 - Pure
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
```python
import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda :sys.stdin.readline().rstrip()
def resolve():
    n,m=map(int,input().split())
    E=[[] for _ in range(n)]
    for _ in range(m):
        a,b=map(lambda x:int(x)-1,input().split())
        E[a].append(b)

    # bfs
    def bfs(sv):
        prev=[-1]*n
        flag=False
        from collections import deque
        Q=deque([sv])
        while(Q):
            v=Q.popleft()
            for nv in E[v]:
                if(nv==sv):
                    flag=True
                    break
                elif(prev[nv]==-1):
                    prev[nv]=v
                    Q.append(nv)
            if(flag): break
        if(not flag):
            return [0]*(n+1)
        else:
            loop=[]
            while(v!=-1):
                loop.append(v)
                v=prev[v]
            loop.reverse()
            return loop

    # calculate
    ans=[0]*(n+1)
    for i in range(n):
        loop=bfs(i)
        if(len(loop)<len(ans)):
            ans=loop
    if(len(ans)==n+1):
        print(-1)
    else:
        print(len(ans))
        for v in ans: print(v+1)
resolve()
```