#include<bits/stdc++.h>
using namespace std;
vector<int>g[100005];
int dis[100005][3];
int main(){
	int n,m,s,t,u,v;
	memset(dis,-1,sizeof(dis));
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
	}
	scanf("%d%d",&s,&t);
	queue<pair<int,int> >q;
	q.push({s,0});
	dis[s][0]=0;
	while(!q.empty()){
		int v=q.front().first;
		int l=q.front().second;
		q.pop();
		for(int i=0;i<g[v].size();i++){
			int ll=(l+1)%3;
			if(dis[g[v][i]][ll]>=0)
				continue;
			dis[g[v][i]][ll]=dis[v][l]+1;
			q.push({g[v][i],ll});
		}
	}
	if(dis[t][0]>=0)
		dis[t][0]/=3;
	printf("%d\n",dis[t][0]);
} 
### Prompt

Please provide a cpp coded solution to the problem described below:

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;
vector<int>g[100005];
int dis[100005][3];
int main(){
	int n,m,s,t,u,v;
	memset(dis,-1,sizeof(dis));
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
	}
	scanf("%d%d",&s,&t);
	queue<pair<int,int> >q;
	q.push({s,0});
	dis[s][0]=0;
	while(!q.empty()){
		int v=q.front().first;
		int l=q.front().second;
		q.pop();
		for(int i=0;i<g[v].size();i++){
			int ll=(l+1)%3;
			if(dis[g[v][i]][ll]>=0)
				continue;
			dis[g[v][i]][ll]=dis[v][l]+1;
			q.push({g[v][i],ll});
		}
	}
	if(dis[t][0]>=0)
		dis[t][0]/=3;
	printf("%d\n",dis[t][0]);
} 
```