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:
/*
	bfs 找环
*/
#include <bits/stdc++.h>
#define N 1005
#define M 2005

using namespace std;

int n, m;
int dep[N];
int head[N], nex[M], to[M], e;
int stk[N], top;
bool instack[N], vis[N], G[N][N];
vector<int> res;

inline int Rd() {
	int x = 0; char ch = getchar();
	while(!isdigit(ch)) ch = getchar();
	while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return x;
}
inline void Ae(int u, int v) {
	nex[++e] = head[u], head[u] = e;
	to[e] = v;
}

void calc(int u, int v) {
	do {
		res.push_back(stk[top]);
	} while(stk[top--] != v);
	int ans = res.size();
	printf("%d\n", ans);
	for(int i = 0; i < ans; ++i) printf("%d\n", res[i]);
	exit(0);
}
void dfs(int u) {
	stk[++top] = u;
	instack[u] = vis[u] = true;
	for(int i = head[u]; i; i = nex[i]) {
		int v = to[i];
		if(vis[v]) {
			if(instack[v]) calc(u, v);
		} else {
			dfs(v);
		}
	}
	instack[u] = false, --top;
	return;
}

int main() {
	n = Rd(), m = Rd();
	for(int i = 1; i <= m; ++i) {
		int u = Rd(), v = Rd();
		G[u][v] = true;
		Ae(u, v);
	}
	
	for(int i = 1; i <= n; ++i) {
		for(int j = 1; j <= n; ++j) {
			if(G[i][j] && G[j][i]) {
				puts("2");
				printf("%d\n%d\n", i, j);
				return 0;
			}	
		}
	}
	for(int i = 1; i <= n; ++i)
		if(!vis[i])
			dfs(i);
	puts("-1");
	return 0;	
}