Problem:
Score : 600 points

Problem Statement

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.

Notes

For a directed graph G = (V, E), we call a directed graph G' = (V', E') satisfying the following conditions an induced subgraph of G:

V' is a (non-empty) subset of V.

E' is the set of all the edges in E that have both endpoints in V'.

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.

Sample Input 1

4 5
1 2
2 3
2 4
4 1
4 3

Sample Output 1

3
1
2
4

The induced subgraph of G whose vertex set is \{1, 2, 4\} has the edge set \{(1, 2), (2, 4), (4, 1)\}. The in-degree and out-degree of every vertex in this graph are both 1.

Sample Input 2

4 5
1 2
2 3
2 4
1 4
4 3

Sample Output 2

-1

There is no induced subgraph of G that satisfies the condition.

Sample Input 3

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2

Sample Output 3

4
2
3
4
5

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

vector < int > with[1003];

vector < int > v;

bool used[1003];
bool inv[1003];

void dfs(int curr) {
	if (inv[curr]) {
		//cout<<"in\n";
		int i = 0;
		for (; v[i] != curr; i++) {}
		cout<<(v.size() - i)<<endl;
		for (int j = i; j < v.size(); j++)
			cout<<v[j]<<endl;
		exit(0);
	}
	if (used[curr]) return;
	used[curr] = 1;
	inv[curr] = 1;
	v.push_back(curr);
	for (int j = 0; j < with[curr].size(); j++) {
		dfs(with[curr][j]);
		if (v.size() > 0) {
			inv[v.back()] = 0;
			v.pop_back();
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	int n, m;
	cin>>n>>m;	
	for (int i = 0; i < m; i++) {
		int l, r;
		cin>>l>>r;
		with[l].push_back(r);
	}
	for (int i = 1; i <= n; i++)
		if (!used[i]) dfs(i);
	cout<<"-1\n";
}