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

int main(){
	int n,m;
	int s,t;
	int flag=0;
	cin>>n>>m;
	vector<vector<int>> graph(3*n);
	int tu,tv;
	for (int i = 0; i < m; i++) {
		cin>>tu>>tv;
		for (int j = 0; j < 3; j++) {
			graph[(tu-1+j*n)%(3*n)].push_back((tv-1+(j+1)*n)%(3*n));
		}
	}
	cin>>s>>t;
	int len=0;
	int dest;
	int temp;
	vector<int> reach(3*n,0);
	queue<int> q;
	q.push(s-1);
	while (len<3*n&&q.size()>0&&flag==0) {
		temp=q.size();
		len++;
		for (int i = 0; i < temp; i++) {
			for (int j  = 0; j  < graph[q.front()].size(); j++) {
				dest=graph[q.front()][j];
				if (reach[dest]==0) {
					q.push(dest);
					reach[dest]=1;
					if (reach[t-1]==1) {
						flag=1;
					}
				}
			}
			q.pop();
		}
	}
	if(flag==0){
		cout<<-1<<endl;
	}else {
		cout<<len/3<<endl;
	}
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph G. G consists of N vertices numbered 1 to N, and M edges. The i-th edge points from Vertex u_i to Vertex v_i.

First, Ken stands on Vertex S. He wants to reach Vertex T by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

Determine if he can reach Vertex T by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex T. Note that visiting Vertex T in the middle of a ken-ken-pa does not count as reaching Vertex T by repeating ken-ken-pa.

Constraints

* 2 \leq N \leq 10^5
* 0 \leq M \leq \min(10^5, N (N-1))
* 1 \leq u_i, v_i \leq N(1 \leq i \leq M)
* u_i \neq v_i (1 \leq i \leq M)
* If i \neq j, (u_i, v_i) \neq (u_j, v_j).
* 1 \leq S, T \leq N
* S \neq T

Input

Input is given from Standard Input in the following format:


N M
u_1 v_1
:
u_M v_M
S T


Output

If Ken cannot reach Vertex T from Vertex S by repeating ken-ken-pa, print -1. If he can, print the minimum number of ken-ken-pa needed to reach vertex T.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2

### Response

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

int main(){
	int n,m;
	int s,t;
	int flag=0;
	cin>>n>>m;
	vector<vector<int>> graph(3*n);
	int tu,tv;
	for (int i = 0; i < m; i++) {
		cin>>tu>>tv;
		for (int j = 0; j < 3; j++) {
			graph[(tu-1+j*n)%(3*n)].push_back((tv-1+(j+1)*n)%(3*n));
		}
	}
	cin>>s>>t;
	int len=0;
	int dest;
	int temp;
	vector<int> reach(3*n,0);
	queue<int> q;
	q.push(s-1);
	while (len<3*n&&q.size()>0&&flag==0) {
		temp=q.size();
		len++;
		for (int i = 0; i < temp; i++) {
			for (int j  = 0; j  < graph[q.front()].size(); j++) {
				dest=graph[q.front()][j];
				if (reach[dest]==0) {
					q.push(dest);
					reach[dest]=1;
					if (reach[t-1]==1) {
						flag=1;
					}
				}
			}
			q.pop();
		}
	}
	if(flag==0){
		cout<<-1<<endl;
	}else {
		cout<<len/3<<endl;
	}
}
```