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 <bits/stdc++.h>
#define debug(x) cerr << #x << " " << (x) << endl
using namespace std;
const int N = 1005, M = 2005;
struct edge {
	int to, nxt;
} graph[M];
int n, m, head[N], par[N], dep[N], cnt = 0, tot = 0;
void addedge (int u, int v) {
	edge x = {v, head[u]};
	graph[head[u] = cnt++] = x;
}
vector<int> ans;
queue<int> q;
void bfs (int s) {
	for (int i = 1; i <= n; i++) dep[i] = -1, par[i] = 0;
	dep[s] = 0, q.push(s);
	while (!q.empty()) {
		int u = q.front();
		q.pop();
		for (int i = head[u]; ~i; i = graph[i].nxt) {
			int v = graph[i].to;
			if (dep[v] < 0) par[v] = u, dep[v] = dep[u] + 1, q.push(v);
		}
	}
}
int main () {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) head[i] = -1;
	for (int i = 1; i <= m; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		addedge(u, v);
	}
	int mn = n + 1, u = 0, v = 0;
	for (int i = 1; i <= n; i++) {
		bfs(i);
		for (int j = 1; j <= n; j++) {
			for (int k = head[j]; ~k; k = graph[k].nxt) {
				int x = graph[k].to;
				if (x == i && dep[j] > 0) {
					if (mn > dep[j] + 1) mn = dep[j] + 1, u = i, v = j;
				}
			}
		}
	}
	if (mn > n) printf("-1\n");
	else {
		bfs(u);
		for (int i = v; i; i = par[i]) ans.push_back(i);
		printf("%d\n", ans.size());
		for (int i = 0; i < ans.size(); i++) printf("%d\n", ans[i]);
	}
	return 0;
}