#include <bits/stdc++.h>
using namespace std;

int n, m;
vector < vector < int > > g;
vector < int > clr, p;
vector < deque < int > > cycles;

void dfs (int u) {
	++clr[u];
	for (int v: g[u]) {
		if (clr[v] == 0) {
			p[v] = u;
			dfs(v);
		}
		else if (clr[v] == 1) {
			deque < int > cycle;
			cycle.push_front(u);
			while (cycle.front() != v) 
				cycle.push_front(p[cycle.front()]);
			cycles.emplace_back(cycle);
		}
	}
	++clr[u];
}

signed main () {
	cin >> n >> m;
	g.resize(n + 1);
	for (int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		g[a].push_back(b);
	}
	for (int root = 1; root <= n; ++root) {
		clr.assign(n + 1, 0);
		p.assign(n + 1, -1);
		dfs(root);
	}
	bool found = 0;
	for (auto cycle: cycles) {
		unordered_set < int > uniq;
		for (int u: cycle) 
			uniq.insert(u);
		bool cap = 0;
		for (int u: cycle) {
			int in_cyc = 0;
			for (int v: g[u]) {
				if (uniq.find(v) != uniq.end())
					++in_cyc;
			}
			if (in_cyc != 1) {
				cap = 1;
				break;
			}
		}
		if (uniq.size() == cycle.size() && !cap) {
			cout << int(cycle.size()) << '\n';
			for (int u: cycle) 
				cout << u << '\n';
			found = 1;
			break;
		}
	}
	if (!found)
		cout << "-1\n";
}

### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int n, m;
vector < vector < int > > g;
vector < int > clr, p;
vector < deque < int > > cycles;

void dfs (int u) {
	++clr[u];
	for (int v: g[u]) {
		if (clr[v] == 0) {
			p[v] = u;
			dfs(v);
		}
		else if (clr[v] == 1) {
			deque < int > cycle;
			cycle.push_front(u);
			while (cycle.front() != v) 
				cycle.push_front(p[cycle.front()]);
			cycles.emplace_back(cycle);
		}
	}
	++clr[u];
}

signed main () {
	cin >> n >> m;
	g.resize(n + 1);
	for (int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		g[a].push_back(b);
	}
	for (int root = 1; root <= n; ++root) {
		clr.assign(n + 1, 0);
		p.assign(n + 1, -1);
		dfs(root);
	}
	bool found = 0;
	for (auto cycle: cycles) {
		unordered_set < int > uniq;
		for (int u: cycle) 
			uniq.insert(u);
		bool cap = 0;
		for (int u: cycle) {
			int in_cyc = 0;
			for (int v: g[u]) {
				if (uniq.find(v) != uniq.end())
					++in_cyc;
			}
			if (in_cyc != 1) {
				cap = 1;
				break;
			}
		}
		if (uniq.size() == cycle.size() && !cap) {
			cout << int(cycle.size()) << '\n';
			for (int u: cycle) 
				cout << u << '\n';
			found = 1;
			break;
		}
	}
	if (!found)
		cout << "-1\n";
}

```