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;

const int MXN = 1005;

vector<int> adj[MXN];

int short_cycle = INT_MAX;
int short_has = 0;
bool vis[MXN];
int st[MXN];

void dfs(int node, int depth, int last) {
	st[node] = depth;
	vis[node] = true;
	for (int i=0; i<adj[node].size(); i++) {
		int other = adj[node][i];
		if (other == last) continue;
		if (st[other] != -1) {
			int x = depth - st[other] + 1;
			if (x < short_cycle) {
				short_cycle = x;
				short_has = node;
			}
		}
		else {
			if (vis[other]) continue;
			dfs(other, depth+1, node);
		}
	}
	st[node] = -1;
}

stack<int> hold;

void dfsit(int node, int depth, int last) {
	st[node] = depth;
	vis[node] = true;
	hold.push(node);
	for (int i=0; i<adj[node].size(); i++) {
		int other = adj[node][i];
		if (other == last) continue;
		if (st[other] != -1) {
			int x = depth - st[other] + 1;
			if (x == short_cycle) {
				while (hold.size() > 0 && hold.top() != other) {
					cout << hold.top() + 1 << endl;
					hold.pop();
				}
				cout << short_has+1 << endl;
				exit(0);
			}
		}
		else {
			// if (vis[other]) continue;
			dfsit(other, depth+1, node);
		}
	}
	hold.pop();
	st[node] = -1;
}

int main() {
	
	int n, m;
	cin >> n >> m;
	for (int i=0; i<m; i++) {
		int a, b;
		cin >> a >> b;
		a--; b--;
		adj[a].push_back(b);
	}
	
	for (int i=0; i<n; i++) st[i] = -1;
	
	dfs(0, 0, -1);
	
	if (short_cycle == INT_MAX) {
		cout << -1 << endl;
		exit(0);
	}
	
	memset(vis, 0, sizeof vis);
	cout << short_cycle << endl;
	dfsit(short_has, 0, -1);
	
}