#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double D;
typedef pair<int,int> P;
#define M 1000000007
#define F first
#define S second
#define PB push_back
#define INF 100000000000000000
int n,m,dp[100005][3],s,t;
vector<int>g[100005];
int main(void){
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int a,b;
		cin>>a>>b;
		g[--a].PB(--b);
	}
	cin>>s>>t;
	s--,t--;
	for(int i=0;i<n;i++){
		dp[i][0]=-1;
		dp[i][1]=-1;
		dp[i][2]=-1;
	}
	dp[s][0]=0;
	queue<P>bfs;
	bfs.push(P(s,0));
	while(!bfs.empty()){
		int v=bfs.front().F,k=bfs.front().S,c=dp[v][k];
		bfs.pop();
		for(int i=0;i<g[v].size();i++){
			int u=g[v][i];
			if(dp[u][(k+1)%3]==-1){
				dp[u][(k+1)%3]=c+1;
				bfs.push(P(u,(k+1)%3));
			}
		}
	}
	if(dp[t][0]!=-1)dp[t][0]/=3;
	cout<<dp[t][0]<<endl;
}
### 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;
typedef long long ll;
typedef double D;
typedef pair<int,int> P;
#define M 1000000007
#define F first
#define S second
#define PB push_back
#define INF 100000000000000000
int n,m,dp[100005][3],s,t;
vector<int>g[100005];
int main(void){
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int a,b;
		cin>>a>>b;
		g[--a].PB(--b);
	}
	cin>>s>>t;
	s--,t--;
	for(int i=0;i<n;i++){
		dp[i][0]=-1;
		dp[i][1]=-1;
		dp[i][2]=-1;
	}
	dp[s][0]=0;
	queue<P>bfs;
	bfs.push(P(s,0));
	while(!bfs.empty()){
		int v=bfs.front().F,k=bfs.front().S,c=dp[v][k];
		bfs.pop();
		for(int i=0;i<g[v].size();i++){
			int u=g[v][i];
			if(dp[u][(k+1)%3]==-1){
				dp[u][(k+1)%3]=c+1;
				bfs.push(P(u,(k+1)%3));
			}
		}
	}
	if(dp[t][0]!=-1)dp[t][0]/=3;
	cout<<dp[t][0]<<endl;
}
```