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 <iostream>
#include <cstdio>
#include <vector>
#define N 1005
using namespace std;

int n, m, mn = 1e9, c[N];
bool v[N];
vector<int> a, ans, gr[N];

void f(int p, int q) {
	int i, t;
	if (c[p]) {
		t = q - c[p];
		if (t < mn) {
			mn = t;
			ans.clear();
			for (i = 1; i <= t; i++) ans.push_back(a[a.size() - i]);
		}
		return;
	}
	if (v[p]) return;
	v[p] = 1;
	c[p] = q;
	a.push_back(p);
	for (i = 0; i < gr[p].size(); i++) {
		f(gr[p][i], q + 1);
	}
	c[p] = 0;
	a.pop_back();
}

int main()
{
	int i, j, t1, t2;
	cin >> n >> m;
	while (m--) {
		scanf("%d %d", &t1, &t2);
		gr[t1].push_back(t2);
	}
	for (i = 1; i <= n; i++) {
		for (j = 1; j <= n; j++) v[j] = 0;
		f(i, 1);
	}
	if (mn == 1e9) {puts("-1"); return 0;}
	cout << mn << endl;
	for (i = 0; i < mn; i++) printf("%d\n", ans[i]);
    return 0;
}