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 rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
typedef vector<int> vi;

int n;
vector<vi> to;

vi state;
vi vs;
bool dfs(int v) {
	if (state[v]) {
		if (state[v] == 1) {
			vs.erase(vs.begin(), find(vs.begin(), vs.end(), v));
			return true;
		}
		return false;
	}
	vs.push_back(v);
	state[v] = 1;
	for (int u : to[v]) {
		if (dfs(u)) return true;
	}
	state[v] = -1;
	vs.pop_back();
	return false;
}

void solve() {
	vector<int> nxt(n), visited(n), erased(n, 1);
	rep(i, vs.size()) {
		int v = vs[i];
		int u = vs[(i + 1) % vs.size()];
		nxt[v] = u;
		erased[v] = 0;
	}
	int v = vs[0];
	while (!visited[v]) {
		for (int u : to[v]) {
			if (erased[u]) continue;
			int nv = v;
			while (nxt[nv] != u) {
				nv = nxt[nv];
				erased[nv] = 1;
			}
			nxt[v] = u;
		}
		visited[v] = 1;
		v = nxt[v];
	}
	vi ans;
	int sv = v;
	while (1) {
		ans.push_back(v);
		v = nxt[v];
		if (v == sv) break;
	}
	cout << ans.size() << endl;
	for (int v : ans) cout << v + 1 << endl;
}

int main() {
	int m;
	cin >> n >> m;
	to.resize(n);
	rep(i, m) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		to[a].push_back(b);
	}

	state = vi(n);
	rep(i, n) if (!state[i]) {
		if (dfs(i)) {
			solve();
			return 0;
		}
	}
	cout << -1 << endl;
	return 0;
}