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

int n,m;


int main(){
	cin >> n >> m;
	vector<vector<int> > g(n*3);
	int f[n*3];
	for(int i =0;i<n*3;++i) f[i] = -1;
	
	int u,v;
	for(int i=0;i<m;++i){
		cin >> u >> v;
		u--;
		v--;
		g[u].push_back(v+n);
		g[u+n].push_back(v+n*2);
		g[u+n*2].push_back(v);
	}
	
	int s,t;
	cin >> s >> t;
	s--;t--;
	s = s + n*2;
	queue<int> q;
	q.push(s);
	f[s] = 0;
	while(!q.empty()){
		int c = q.front();q.pop();
		for(int i=0;i<g[c].size();++i){
			int next = g[c][i];
//			cout << next << endl;
			if(f[next] == -1){
				q.push(next);
				f[next] = f[c] + 1;
			}
		}
	}
	
	if(f[t+2*n] == -1) cout << -1 << endl;
	else cout << f[t+2*n]/3 << endl;
	/*
	for(int i =0;i<n;++i){
		cout << f[i] << " " << f[i+n] << " " << f[i+2*n] <<endl;
	}*/

	return 0;
}


### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#include<queue>
using namespace std;

int n,m;


int main(){
	cin >> n >> m;
	vector<vector<int> > g(n*3);
	int f[n*3];
	for(int i =0;i<n*3;++i) f[i] = -1;
	
	int u,v;
	for(int i=0;i<m;++i){
		cin >> u >> v;
		u--;
		v--;
		g[u].push_back(v+n);
		g[u+n].push_back(v+n*2);
		g[u+n*2].push_back(v);
	}
	
	int s,t;
	cin >> s >> t;
	s--;t--;
	s = s + n*2;
	queue<int> q;
	q.push(s);
	f[s] = 0;
	while(!q.empty()){
		int c = q.front();q.pop();
		for(int i=0;i<g[c].size();++i){
			int next = g[c][i];
//			cout << next << endl;
			if(f[next] == -1){
				q.push(next);
				f[next] = f[c] + 1;
			}
		}
	}
	
	if(f[t+2*n] == -1) cout << -1 << endl;
	else cout << f[t+2*n]/3 << endl;
	/*
	for(int i =0;i<n;++i){
		cout << f[i] << " " << f[i+n] << " " << f[i+2*n] <<endl;
	}*/

	return 0;
}


```