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 IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << #x << " is " << x << endl

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;

const int INF = 1e9 + 7;

vi G[1005], s;
bool ok[1005], vis[1005];

void dfs(int u, int tar) {
	s.PB(u);
	vis[u] = 1;
	if(u == tar) {
		cout << s.size() << endl;
		for(int e : s) cout << e << endl;
		exit(0);
	}
	for(int v:G[u]) if(ok[v])
		dfs(v, tar);
	s.pop_back();
}

int32_t main()
{
	IO_OP;
	
	int n, m;
	cin >> n >> m;
	for(int i=0;i<m;i++) {
		int u, v;
		cin >> u >> v;
		G[u].PB(v);
	}
	for(int i=1;i<=n;i++) {
		vi undo;
		for(int u:G[i]) {
			if(u < i) {
				undo.PB(u);
				ok[u] = false;
			}
		}
		memset(vis, 0, sizeof vis);
		ok[i] = 1;
		for(int u:undo) {
			ok[u] = 1;
			dfs(u, i);
			ok[u] = 0;
		}
		for(int u:undo) ok[u] = 1;
	}
	cout << -1 << endl;
}