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:
// これは嘘解法です！！

#include <bits/stdc++.h>
using namespace std;
#define inc(i, n) for(int i = 0; i < (n); i++)

int n, m;
vector<int> g[1000];
bool vis[1000];

bool dfs(int v, int s) {
	if(vis[v]) { return false; }
	vis[v] = true;
	for(auto & e: g[v]) {
		if(e == s || dfs(e, s)) { return true; }
	}
	return false;
}

bool f(int v) {
	inc(i, n) { vis[i] = false; }
	return dfs(v, v);
}

int main() {
	cin >> n >> m;
	inc(i, m) {
		int a, b;
		cin >> a >> b;
		a--; b--;
		g[a].push_back(b);
	}
	
	bool ans = false;
	inc(i, n) { ans |= f(i); }
	
	cout << (ans ? 0 : -1) << endl;
	
	return 0;
}