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;
const int MAXN=2010;
int n,m,x,y,eg,len,pa,pb,hd[2][MAXN],ver[2][MAXN*2],nx[2][MAXN*2],vis[MAXN],dis[MAXN],las[MAXN];
void add_edge (int t,int x,int y) {
	ver[t][++eg]=y;
	nx[t][eg]=hd[t][x];
	hd[t][x]=eg;
	return;
}
void dij (int x) {
	memset(dis,0x3f,sizeof(dis));
	memset(vis,0,sizeof(vis));
	memset(las,0,sizeof(las));
	dis[x]=0,vis[x]=1;
	queue <int> q;
	q.push(x);
	while (!q.empty()) {
		int a=q.front();
		q.pop();
		for (int i=hd[0][a];i;i=nx[0][i]) {
			if (!vis[ver[0][i]]) {
				dis[ver[0][i]]=dis[a]+1,las[ver[0][i]]=a;
				vis[ver[0][i]]=1;
				q.push(ver[0][i]);
			}
		}
	}
	return;
}
int main () {
	scanf("%d%d",&n,&m);
	len=n+1;
	for (int i=1;i<=m;i++) {
		scanf("%d%d",&x,&y);
		add_edge(0,x,y),add_edge(1,y,x);
	}
	for (int i=1;i<=n;i++) {
		dij(i);
		for (int j=hd[1][i];j;j=nx[1][j]) {
			if (dis[ver[1][j]]+1<len) {
				len=dis[ver[1][j]]+1,pa=i,pb=ver[1][j];
			}
		}
	}
	if (len==n+1) {
		printf("-1\n");
	}
	else {
		dij(pa);
		printf("%d\n%d\n",len,pa);
		while (pb!=pa) {
			printf("%d\n",pb);
			pb=las[pb];
		}
	}
	return 0;
}