#include<bits/stdc++.h>
using namespace std;
const int M=1e5+5;
struct node{
	int x,step,mod;
};
bool mark[M][3];
vector<int>Edge[M];
int ans;

bool bfs(int st, int ed){
	queue<node>Q;
	while(!Q.empty()) Q.pop();
	Q.push((node)<%st,0,0%>);
	while(!Q.empty()){
		node now=Q.front(); Q.pop();
		for(int i=0,_i=Edge[now.x].size(); i<_i; i++){
			int u=Edge[now.x][i],v=(now.mod+1)%3;
			if(mark[u][v]) continue;
			mark[u][v]=1;
			if(u==ed && v==0){
				ans=(now.step+1)/3;
				return 1;
			}
			Q.push((node)<%u,now.step+1,v%>);
		}
	}
	return 0;
}

int main(){
	int n,m,st,ed;
	scanf("%d %d",&n,&m);
	for(int i=1,x,y; i<=m; i++){
		scanf("%d %d",&x,&y);
		Edge[x].push_back(y);
	}
	scanf("%d %d",&st,&ed);
	if(bfs(st,ed)) printf("%d\n",ans);
	else puts("-1");
	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 M=1e5+5;
struct node{
	int x,step,mod;
};
bool mark[M][3];
vector<int>Edge[M];
int ans;

bool bfs(int st, int ed){
	queue<node>Q;
	while(!Q.empty()) Q.pop();
	Q.push((node)<%st,0,0%>);
	while(!Q.empty()){
		node now=Q.front(); Q.pop();
		for(int i=0,_i=Edge[now.x].size(); i<_i; i++){
			int u=Edge[now.x][i],v=(now.mod+1)%3;
			if(mark[u][v]) continue;
			mark[u][v]=1;
			if(u==ed && v==0){
				ans=(now.step+1)/3;
				return 1;
			}
			Q.push((node)<%u,now.step+1,v%>);
		}
	}
	return 0;
}

int main(){
	int n,m,st,ed;
	scanf("%d %d",&n,&m);
	for(int i=1,x,y; i<=m; i++){
		scanf("%d %d",&x,&y);
		Edge[x].push_back(y);
	}
	scanf("%d %d",&st,&ed);
	if(bfs(st,ed)) printf("%d\n",ans);
	else puts("-1");
	return 0;
}
```