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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> d[1005],a;
int v,e;
bool c[1005],dd[1005][1005];
bool f(int n,int m){
    if(c[m])return m==n;
    c[m]=true;
    for(int i : d[m]){
        a.push_back(i);
        if(f(n,i))return true;
        a.pop_back();
    }
    return false;
}
int main(){
    scanf("%d %d",&v,&e);
    for(int i=0;i<e;i++){
        int x,y;
        scanf("%d %d",&x,&y);
        d[x].push_back(y);
        dd[x][y]=true;
    }
    for(int i=1;i<=v;i++){
        memset(c,0,v+1);
        a.push_back(i);
        if(f(i,i)){
            int x=0,y=a.size()-2,m=a.size()-1;
            for(int j=0;j<a.size()-1;j++)for(int k=0;k<a.size()-1;k++){
                if(dd[a[k]][a[j]])if(((k-j+1+a.size()-1)%(a.size()-1))<m&&(k-j+1+a.size()-1)%(a.size()-1)){
                    m=(k-j+1+a.size()-1)%(a.size()-1);
                    x=j,y=k;
                }
            }
            printf("%d\n",m);
            for(int j=0;j<m;j++)printf("%d\n",a[(x+j)%(a.size()-1)]);
            return 0;
        }
        a.pop_back();
    }
    printf("-1");
}