
#include <iostream>
#include<vector>
using namespace std;

int main() {
	int n,m;cin>>n>>m;
	vector<int> u(m),v(m);
	vector<vector<int>> uv(n);
	for(int i=0;i<m;i++){
		cin>>u[i]>>v[i];
		u[i]--;v[i]--;
		uv[u[i]].push_back(v[i]);
	}
	int s,t;cin>>s>>t;
	s--;t--;
	int hop=0;
	vector<vector<int>> used(n,vector<int>(3,0));
	vector<int> tempn;
	tempn.push_back(s);
	used[s][0]=1;
	while(tempn.size()){
		vector<int> nextn;
		hop++;
		for(int i=0;i<tempn.size();i++){
			for(int j=0;j<uv[tempn[i]].size();j++){
				if(used[uv[tempn[i]][j]][hop%3]==0){
					nextn.push_back(uv[tempn[i]][j]);
					used[uv[tempn[i]][j]][hop%3]=1;
					if(uv[tempn[i]][j]==t){
						if(hop%3==0){
							cout<<hop/3<<endl;
							return 0;
						}
					}
				}
			}

		}
		tempn=nextn;

	}
	cout<<-1<<endl;


	return 0;
}
### Prompt

Generate 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 <iostream>
#include<vector>
using namespace std;

int main() {
	int n,m;cin>>n>>m;
	vector<int> u(m),v(m);
	vector<vector<int>> uv(n);
	for(int i=0;i<m;i++){
		cin>>u[i]>>v[i];
		u[i]--;v[i]--;
		uv[u[i]].push_back(v[i]);
	}
	int s,t;cin>>s>>t;
	s--;t--;
	int hop=0;
	vector<vector<int>> used(n,vector<int>(3,0));
	vector<int> tempn;
	tempn.push_back(s);
	used[s][0]=1;
	while(tempn.size()){
		vector<int> nextn;
		hop++;
		for(int i=0;i<tempn.size();i++){
			for(int j=0;j<uv[tempn[i]].size();j++){
				if(used[uv[tempn[i]][j]][hop%3]==0){
					nextn.push_back(uv[tempn[i]][j]);
					used[uv[tempn[i]][j]][hop%3]=1;
					if(uv[tempn[i]][j]==t){
						if(hop%3==0){
							cout<<hop/3<<endl;
							return 0;
						}
					}
				}
			}

		}
		tempn=nextn;

	}
	cout<<-1<<endl;


	return 0;
}
```