#include <bits/stdc++.h>

using namespace std;
const int N=1e3+10;


int from[N*2], to[N*2];
int n, m;
vector<int> V[N];
int vis[N], F[N];
int ans;
map<int, map<int, int>> p;

int bfs(int st) {
	queue<int> Q;
	memset(vis, -1, sizeof(vis));
	memset(F, 0, sizeof(F));
	Q.push(st);
	vis[st]=1;
	F[st]=0;
	while(!Q.empty()) {
		int cur=Q.front();Q.pop();
		for(auto k:V[cur]) {
			int u=to[k];
			if (u==st) return ans=cur;
			if (vis[u]!=-1) continue;
			vis[u]=vis[cur]+1;
			F[u]=cur;
			Q.push(u);
		}
	}
	return ans=-1;
}

int main() {
	cin>>n>>m;
	for(int i=1; i<=m; i++) {
		cin>>from[i]>>to[i];
		V[from[i]].push_back(i);
	}
	int len=12345678;
	int A=0;
	for(int i=1; i<=n; i++) {
		int tmp=bfs(i);
		if (tmp==-1) continue;
		if (vis[tmp]<len) {
			A=i;
			len=vis[tmp];
		}
	}
	if (!A) {
		cout<<-1<<endl;
		return 0;
	}
	bfs(A);
	cout<<len<<endl;
	for(int i=1; i<=len; i++) {
		cout<<ans<<endl;
		ans=F[ans];
	}
	return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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=1e3+10;


int from[N*2], to[N*2];
int n, m;
vector<int> V[N];
int vis[N], F[N];
int ans;
map<int, map<int, int>> p;

int bfs(int st) {
	queue<int> Q;
	memset(vis, -1, sizeof(vis));
	memset(F, 0, sizeof(F));
	Q.push(st);
	vis[st]=1;
	F[st]=0;
	while(!Q.empty()) {
		int cur=Q.front();Q.pop();
		for(auto k:V[cur]) {
			int u=to[k];
			if (u==st) return ans=cur;
			if (vis[u]!=-1) continue;
			vis[u]=vis[cur]+1;
			F[u]=cur;
			Q.push(u);
		}
	}
	return ans=-1;
}

int main() {
	cin>>n>>m;
	for(int i=1; i<=m; i++) {
		cin>>from[i]>>to[i];
		V[from[i]].push_back(i);
	}
	int len=12345678;
	int A=0;
	for(int i=1; i<=n; i++) {
		int tmp=bfs(i);
		if (tmp==-1) continue;
		if (vis[tmp]<len) {
			A=i;
			len=vis[tmp];
		}
	}
	if (!A) {
		cout<<-1<<endl;
		return 0;
	}
	bfs(A);
	cout<<len<<endl;
	for(int i=1; i<=len; i++) {
		cout<<ans<<endl;
		ans=F[ans];
	}
	return 0;
}

```