#include <stdio.h>
#include <algorithm>
#include <tuple>
#include <vector>
#include <queue>
using namespace std;
using ll = long long;
using pii = pair<int, int>;

int n, m;
vector<int> e[1010];

int v[1010], last[1010];
vector<int> r(2000, 0);
void bfs(int node) {
	queue<int> q;
	for (int x : e[node])
		q.emplace(x), v[x] = last[x] = node;
	while (!q.empty()) {
		int cur = q.front();
		q.pop();

		for (int x : e[cur])
			if (v[x] != node)
				q.emplace(x), v[x] = node, last[x] = cur;
	}
	if (v[node] != node)
		return;
	int pop = node;
	vector<int> path;
	path.emplace_back(node);
	while (last[pop] != node) {
		path.emplace_back(last[pop]);
		pop = last[pop];
	}
	if (path.size() < r.size())
		r.swap(path);
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; i++) {
		int a, b;
		scanf("%d %d", &a, &b);
		e[a].emplace_back(b);
	}
	for (int i = 1; i <= n; i++)
		bfs(i);
	if (n < r.size())
		printf("-1\n");
	else {
		printf("%d\n", r.size());
		for (int v : r)
			printf("%d\n", v);
	}
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <stdio.h>
#include <algorithm>
#include <tuple>
#include <vector>
#include <queue>
using namespace std;
using ll = long long;
using pii = pair<int, int>;

int n, m;
vector<int> e[1010];

int v[1010], last[1010];
vector<int> r(2000, 0);
void bfs(int node) {
	queue<int> q;
	for (int x : e[node])
		q.emplace(x), v[x] = last[x] = node;
	while (!q.empty()) {
		int cur = q.front();
		q.pop();

		for (int x : e[cur])
			if (v[x] != node)
				q.emplace(x), v[x] = node, last[x] = cur;
	}
	if (v[node] != node)
		return;
	int pop = node;
	vector<int> path;
	path.emplace_back(node);
	while (last[pop] != node) {
		path.emplace_back(last[pop]);
		pop = last[pop];
	}
	if (path.size() < r.size())
		r.swap(path);
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; i++) {
		int a, b;
		scanf("%d %d", &a, &b);
		e[a].emplace_back(b);
	}
	for (int i = 1; i <= n; i++)
		bfs(i);
	if (n < r.size())
		printf("-1\n");
	else {
		printf("%d\n", r.size());
		for (int v : r)
			printf("%d\n", v);
	}
}
```