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 <queue>
#include <cstdio>
#include <algorithm>
#include <string>
#include <vector>
#define FOR(i, n) for(int i = 0; i < n; i++) 
using namespace std;

typedef pair<int,int> pii;
typedef long long ll;

void load() {
}

const int MAXN = 1010;
int n, m;
vector<int> graf[MAXN];

vector<int> bfs(int s) {
	queue<int> q;
	vector<int> parent(n, -1);
	q.push(s);
	while (!q.empty()) {
		int x = q.front(); q.pop();
		for (int sus : graf[x]) {
			if (parent[sus] == -1) {
				parent[sus] = x;
				if (sus == s) {
					vector<int> sol;
					do {
						sol.push_back(sus);
						sus = parent[sus];
					} while (sus != s);
					return sol;
				}
				q.push(sus);
			}
		}
	}
	vector<int> v;
	return v;
}

void solve() {
	scanf("%d%d", &n, &m);
	FOR(i, m) {
		int u, v; scanf("%d%d", &u, &v); u--; v--;
		graf[u].push_back(v);
	}
	vector<int> sol;
	FOR(i, n) {
		vector<int> v = bfs(i);
		if (v.size() > 0 && (sol.empty() || v.size() < sol.size())) sol = v;
	}
	if (sol.empty()) {
		printf("-1\n");
	} else {
		printf("%d\n", (int)sol.size());
		for (int y : sol) printf("%d\n", y+1);
	}
}

int main() {
	load();
	solve();
	return 0;
}