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(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda:sys.stdin.readline().rstrip()
from collections import deque
def resolve():
    n,m=map(int,input().split())
    E=[[] for _ in range(n)]
    for _ in range(m):
        u,v=map(int,input().split())
        u-=1; v-=1
        E[u].append(v)

    # 各頂点から BFS を行い、全点間の距離をO(NM)で計算
    A=[None]*n
    for s in range(n):
        dist=[INF]*n
        dist[s]=0
        Q=deque([s])
        while(Q):
            v=Q.popleft()
            for nv in E[v]:
                if(dist[nv]!=INF): continue
                dist[nv]=dist[v]+1
                if(nv!=s): Q.append(nv)
            if(v==s): dist[v]=INF
        A[s]=dist

    d,v0=min((A[v][v],v) for v in range(n))
    if(d==INF):
        print(-1)
        return

    print(d)
    # v から v の最小のループを求める
    prev=[None]*n
    used=[0]*n
    used[0]=1
    Q=deque([v0])
    while(Q):
        flag=True
        v=Q.popleft()
        for nv in E[v]:
            if(nv==v0):
                flag=False
                break
            if(used[nv]): continue
            used[nv]=1
            prev[nv]=v
            Q.append(nv)
        if(not flag): break

    ans=[]
    while(v!=v0):
        ans.append(v)
        v=prev[v]

    ans.append(v0)
    print(*map(lambda x:x+1,ans[::-1]),sep='\n')
resolve()