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

const int maxn=1e5+100;

queue <pair<int,int> > Q;
bool vis[maxn][3];
vector <int> g[maxn];
int n,m;
int st,ed;

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		int fr,to;
		scanf("%d%d",&fr,&to);
		g[fr].push_back(to);
	}
	scanf("%d%d",&st,&ed);
	Q.push(make_pair(st,0));
	vis[st][0]=1;
	while(!Q.empty()){
		int x=Q.front().first;
		int h=Q.front().second;
		Q.pop();
		if(x==ed&&h%3==0){
			cout<<h/3<<endl;
			return 0;
		}
		h++;
		for(int i=0;i<g[x].size();i++){	
			int to=g[x][i];
			if(!vis[to][h%3]){
				Q.push(make_pair(to,h));
				vis[to][h%3]=1;
			}
		}
	}
	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;

const int maxn=1e5+100;

queue <pair<int,int> > Q;
bool vis[maxn][3];
vector <int> g[maxn];
int n,m;
int st,ed;

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		int fr,to;
		scanf("%d%d",&fr,&to);
		g[fr].push_back(to);
	}
	scanf("%d%d",&st,&ed);
	Q.push(make_pair(st,0));
	vis[st][0]=1;
	while(!Q.empty()){
		int x=Q.front().first;
		int h=Q.front().second;
		Q.pop();
		if(x==ed&&h%3==0){
			cout<<h/3<<endl;
			return 0;
		}
		h++;
		for(int i=0;i<g[x].size();i++){	
			int to=g[x][i];
			if(!vis[to][h%3]){
				Q.push(make_pair(to,h));
				vis[to][h%3]=1;
			}
		}
	}
	cout<<"-1"<<endl;
	return 0;
}
```