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<cstdio>
#include<vector>
using namespace std;
const int MAXN = 1000;
vector<int>G[MAXN + 5];
bool tag[MAXN + 5], vis[MAXN + 5], flag;
int stk[MAXN + 5], tp, val;
void dfs(int x) {
	stk[++tp] = x, vis[x] = tag[x] = true;
	for(int i=0;i<G[x].size();i++) {
		int p = G[x][i];
		if( !vis[p] ) dfs(p);
		else if( tag[p] )
			val = p, flag = true;
		if( flag ) return ;
	}
	stk[tp--] = 0, vis[x] = false;
}
int nxt[MAXN + 5];
vector<int>ans;
int main() {
	int N, M; scanf("%d%d", &N, &M);
	for(int i=1;i<=M;i++) {
		int A, B; scanf("%d%d", &A, &B);
		G[A].push_back(B);
	}
	for(int i=1;i<=N;i++)
		if( !tag[i] ) {
			dfs(i);
			if( flag ) break;
		}
	if( !flag ) {
		puts("-1");
		return 0;
	}
	int pos;
	for(int i=1;i<=tp;i++)
		if( stk[i] == val ) pos = i;
	nxt[stk[tp]] = val;
	for(int i=pos;i<tp;i++)
		nxt[stk[i]] = stk[i+1];
	for(int i=1;i<=N;i++) {
		if( !nxt[i] ) continue;
		for(int j=0;j<G[i].size();j++) {
			int x = G[i][j];
			if( !nxt[x] ) continue;
			int p = i, q = nxt[i];
			while( q != x ) {
				int r = nxt[q];
				nxt[p] = r;
				nxt[q] = 0;
				q = r;
			}
		}
	}
	int x;
	for(int i=1;i<=N;i++)
		if( nxt[i] ) x = i;
	int p = x;
	do {
		ans.push_back(p);
		p = nxt[p];
	}while( p != x );
	printf("%d\n", ans.size());
	for(int i=0;i<ans.size();i++)
		printf("%d\n", ans[i]);
}