#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;

#define MAX 100000
#define INF -1

int n;
vector<int> g[4*MAX];
int d[4*MAX];

void bfs(int s){
	queue<int> q;
	fill(d, d+4*MAX, INF);
	q.push(s);
	d[s]=0;

	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=0; i<g[u].size(); i++){
			if(d[g[u][i]]==INF){
				q.push(g[u][i]);
				d[g[u][i]]=d[u]+1;
			}
		}
	}
	return;
}

int main(){
	int m;
	cin >> n >> m;
	for(int i=0; i<m; i++){
		int u, v;
		cin >> u >> v;
		u--; v--;
		for(int j=0; j<3; j++){
			g[j*MAX+u].push_back(((j+1)%3)*MAX+v);
		}
	}

	int s, t;
	cin >> s >> t;
	s--; t--;

	bfs(s);
	if(d[t]!=INF && d[t]%3==0) cout << d[t]/3 << endl;
	else cout << -1 << 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 <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;

#define MAX 100000
#define INF -1

int n;
vector<int> g[4*MAX];
int d[4*MAX];

void bfs(int s){
	queue<int> q;
	fill(d, d+4*MAX, INF);
	q.push(s);
	d[s]=0;

	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=0; i<g[u].size(); i++){
			if(d[g[u][i]]==INF){
				q.push(g[u][i]);
				d[g[u][i]]=d[u]+1;
			}
		}
	}
	return;
}

int main(){
	int m;
	cin >> n >> m;
	for(int i=0; i<m; i++){
		int u, v;
		cin >> u >> v;
		u--; v--;
		for(int j=0; j<3; j++){
			g[j*MAX+u].push_back(((j+1)%3)*MAX+v);
		}
	}

	int s, t;
	cin >> s >> t;
	s--; t--;

	bfs(s);
	if(d[t]!=INF && d[t]%3==0) cout << d[t]/3 << endl;
	else cout << -1 << endl;
	return 0;
}
```