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 <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;

int N, M, D[1010], C[1010];
vector<int> G[1010];

void go(int x, int d)
{
	D[x] = d;

	int v = 0;
	for (auto &y : G[x]){
		C[y]++;
		v = max(v, D[y]);
	}

	if (v){
		d = 0;
		for (int i = 1; i <= N; i++) if (D[i] >= v) d++;
		printf ("%d\n", d);
		for (int i = 1; i <= N; i++) if (D[i] >= v) printf ("%d\n", i);
		exit(0);
	}

	for (auto &y : G[x]){
		C[y]--;
		if (C[y] == 0) go(y, d + 1);
		C[y]++;
	}
	for (auto &y : G[x]) C[y]--;

	D[x] = 0;
}

int main()
{
	scanf ("%d %d", &N, &M);
	for (int i = 1; i <= M; i++){
		int x, y; scanf ("%d %d", &x, &y);
		G[x].push_back(y);
	}

	for (int i = 1; i <= N; i++) go(i, 1);
	puts("-1");
	return 0;
}