# include <bits/stdc++.h>
using namespace std;
const int N = 1e5 * 3 + 10;
vector < int > edges[N];
int dist[N] , fix[N];
int main(){
	int n , m;
	cin >> n >> m;
	for(int i = 1; i <= m; i ++){
		int u , v;
		scanf("%d%d" , &u , &v);
		edges[u].push_back(v + n);
		edges[u + n].push_back(v + 2 * n);
		edges[u + 2 * n].push_back(v);
	}
	int S , T;
	cin >> S >> T;
	queue < pair < int , int > > Q;
	Q.push({S , 0});
	while(Q.size()){
		int u = Q.front().first;
		int d = Q.front().second;
		Q.pop();
		if(fix[u]) continue;
		fix[u] = 1;
		dist[u] = d;
		if(u == T) break;
		for(int i = 0; i < edges[u].size(); i ++)
			Q.push({edges[u][i] , d + 1});	
	}
	cout << (fix[T] ? dist[T] / 3 : -1) << "\n";
}
### 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 <bits/stdc++.h>
using namespace std;
const int N = 1e5 * 3 + 10;
vector < int > edges[N];
int dist[N] , fix[N];
int main(){
	int n , m;
	cin >> n >> m;
	for(int i = 1; i <= m; i ++){
		int u , v;
		scanf("%d%d" , &u , &v);
		edges[u].push_back(v + n);
		edges[u + n].push_back(v + 2 * n);
		edges[u + 2 * n].push_back(v);
	}
	int S , T;
	cin >> S >> T;
	queue < pair < int , int > > Q;
	Q.push({S , 0});
	while(Q.size()){
		int u = Q.front().first;
		int d = Q.front().second;
		Q.pop();
		if(fix[u]) continue;
		fix[u] = 1;
		dist[u] = d;
		if(u == T) break;
		for(int i = 0; i < edges[u].size(); i ++)
			Q.push({edges[u][i] , d + 1});	
	}
	cout << (fix[T] ? dist[T] / 3 : -1) << "\n";
}
```