#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
struct edge {int to, next;} e[N * 2];
int stk[N], top, vis[N];
int head[N], ecnt, rt;
inline void adde(int from, int to) {e[++ecnt] = (edge){to, head[from]}, head[from] = ecnt;}
int ans[N], fuck = 1e9;
void dfs(int u) {
	if (top >= fuck) return;
	if (vis[u]) {
		if (u != rt) return;
		if (top < fuck) {
			fuck = top;
			for (int i = 1; i <= top; i++) ans[i] = stk[i];
		}
		return;
	}
	vis[u] = 1, stk[++top] = u;
	for (int i = head[u]; i; i = e[i].next)
		dfs(e[i].to);
	top--;
}
int main() {
	int n, m; scanf("%d%d", &n, &m);
	for (int i = 1, a, b; i <= m; i++) scanf("%d%d", &a, &b), adde(a, b);
	for (int i = 1; i <= n; i++) {
		memset(vis, 0, sizeof(vis));
		rt = i, dfs(i);
	}
	if (fuck < 1e9) {
		cout << fuck << endl;
		for (int i = 1; i <= fuck; i++) cout << ans[i] << endl;
		return 0; 
	}
	puts("-1");
}
### Prompt

In CPP, your task is to solve 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;
const int N = 1010;
struct edge {int to, next;} e[N * 2];
int stk[N], top, vis[N];
int head[N], ecnt, rt;
inline void adde(int from, int to) {e[++ecnt] = (edge){to, head[from]}, head[from] = ecnt;}
int ans[N], fuck = 1e9;
void dfs(int u) {
	if (top >= fuck) return;
	if (vis[u]) {
		if (u != rt) return;
		if (top < fuck) {
			fuck = top;
			for (int i = 1; i <= top; i++) ans[i] = stk[i];
		}
		return;
	}
	vis[u] = 1, stk[++top] = u;
	for (int i = head[u]; i; i = e[i].next)
		dfs(e[i].to);
	top--;
}
int main() {
	int n, m; scanf("%d%d", &n, &m);
	for (int i = 1, a, b; i <= m; i++) scanf("%d%d", &a, &b), adde(a, b);
	for (int i = 1; i <= n; i++) {
		memset(vis, 0, sizeof(vis));
		rt = i, dfs(i);
	}
	if (fuck < 1e9) {
		cout << fuck << endl;
		for (int i = 1; i <= fuck; i++) cout << ans[i] << endl;
		return 0; 
	}
	puts("-1");
}
```