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

const int mn=3e5+7;
int dis[mn];
vector<int> to[mn];

queue<int> q;
void bfs(int s)
{
	dis[s]=1;
	q.push(s);
	while(!q.empty()) {
		s=q.front();q.pop();
		for(int i=0;i<to[s].size();++i) {
			int t=to[s][i];
			if(dis[t]) continue;
			dis[t]=dis[s]+1;q.push(t);
		}
	}
}

int main()
{
	int n,m,s,t,u,v;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;++i) {scanf("%d%d",&u,&v);
		to[u*3].pb(v*3+1),to[u*3+1].pb(v*3+2),to[u*3+2].pb(v*3);
	}
	scanf("%d%d",&s,&t);
	bfs(s*3);
	printf("%d",((dis[t*3]-dis[s*3])%3==0)?(dis[t*3]-dis[s*3])/3:-1);
}
### Prompt

Create a solution in Cpp for 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>
#define pb push_back
using namespace std;

const int mn=3e5+7;
int dis[mn];
vector<int> to[mn];

queue<int> q;
void bfs(int s)
{
	dis[s]=1;
	q.push(s);
	while(!q.empty()) {
		s=q.front();q.pop();
		for(int i=0;i<to[s].size();++i) {
			int t=to[s][i];
			if(dis[t]) continue;
			dis[t]=dis[s]+1;q.push(t);
		}
	}
}

int main()
{
	int n,m,s,t,u,v;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;++i) {scanf("%d%d",&u,&v);
		to[u*3].pb(v*3+1),to[u*3+1].pb(v*3+2),to[u*3+2].pb(v*3);
	}
	scanf("%d%d",&s,&t);
	bfs(s*3);
	printf("%d",((dis[t*3]-dis[s*3])%3==0)?(dis[t*3]-dis[s*3])/3:-1);
}
```