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 <stdio.h>
#include <algorithm>
#include <string.h>
#include <vector>
#include <stdlib.h>
#include <time.h>
using namespace std;
int n, m, tmpx, tmpy, r[1005], st[1005], ans[1005];
vector< int > edg[1005];
int dfs(int nown) {
	if(st[nown]) return 1;
	st[nown]= 1;
	for(int i= 0; i < edg[nown].size(); i++) {
		++r[edg[nown][i]];
		if(r[edg[nown][i]] > 1 && st[edg[nown][i]]) {
			for(int j= 0; j <= i; j++) --r[edg[nown][j]];
			st[nown]= 0;
			return 0;
		}
	}
	for(int i= 0; i < edg[nown].size(); i++) {
		if(r[edg[nown][i]] == 1 && dfs(edg[nown][i])) {
			ans[++ans[0]]= nown;
			return 1;
		}
	}
	for(int i= 0; i < edg[nown].size(); i++) --r[edg[nown][i]];
	st[nown]= 0;
	return 0;
}
int main() {
    srand(time(0));
	cin >> n >> m;
	for(int i= 1; i <= m; i++) {
		cin >> tmpx >> tmpy;
		edg[tmpx].push_back(tmpy);
	}
	while(clock() < CLOCKS_PER_SEC * 1.8)
		if(dfs(rand() % n + 1)) break;
	if(ans[0] == 0)
		cout << -1 << endl;
	else {
		cout << ans[0] << endl;
		sort(ans + 1, ans + ans[0] + 1);
		for(int i= 1; i <= ans[0]; i++) cout << ans[i] << endl;
	}
	return 0;
}