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: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdlib>
#include <map>
#include <iomanip>
#include <set>
#include <stdio.h>
#include <ctype.h>
#include <random>
#include <string.h>
using namespace std;
#define pie 3.141592653589793238462643383279
#define mod 1000000007
#define all(vec) vec.begin(),vec.end()

int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x%y);
}
int lcm(int x, int y) {
	return x / gcd(x, y)*y;
}
int kai(int x, int y) {
	int res = 1;
	for (int i = x - y + 1; i <= x; i++) {
		res *= i; res %= mod;
	}
	return res;
}
int mod_pow(int x, int y, int m) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % m;
		}
		x = x * x % m;
		y >>= 1;
	}
	return res;
}

int comb(int x, int y) {
	if (y > x)return 0;
	return kai(x, y) * mod_pow(kai(y, y), mod - 2, mod) % mod;
}

int n, m;
vector<int> vec[1010];
bool bl[1010];
bool g[1010][1010];
vector<int> ve;
vector<int> ans;

bool dfs(int v, int p) {
	bl[v] = true;
	for (int i = 0; i < (int)vec[v].size(); i++) {
		if (vec[v][i] == p) {
			ve.push_back(v);
			return true;
		}
		if (bl[vec[v][i]])continue;
		if (dfs(vec[v][i], p)) {
			ve.push_back(v);
			return true;
		}
	}
	return false;
}

signed main() {
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b; cin >> a >> b;
		vec[a].push_back(b);
		g[a][b] = true;
	}
	for (int i = 1; i <= n; i++) {
		fill(bl, bl + n + 1, false);
		if (dfs(i, i)) break;
	}
	if (ve.empty()) {
		cout << -1 << endl;
		return 0;
	}
	reverse(all(ve));
	int s = 0;
	bool bl2 = false;
	for (int i = 0; i < (int)ve.size(); i++) {
		for (int j = (int)ans.size() - 1; j >= 0; j--) {
			if (g[ve[i]][ans[j]]) {
				s = j;
				bl2 = true;
				break;
			}
		}
		ans.push_back(ve[i]);
		if (bl2)break;
		int k = i + 1;
		for (int j = i + 1; j < (int)ve.size(); j++) {
			if (g[ve[i]][ve[j]])k = j;
		}
		i = k - 1;
	}
	cout << (int)ans.size() - s << endl;
	for (int i = s; i < (int)ans.size(); i++)cout << ans[i] << endl;
}