#include<bits/stdc++.h>
#define M 100002
using namespace std;
struct lll{
	int l,to;
}edge[M];
struct ll{
	int x,flg,step;
};
queue<ll>Q;
int h[M],S,T;
bool mem[5][M];
void Add(int i,int st,int to){
	edge[i].l=h[st];
	edge[i].to=to;
	h[st]=i;
}
int bfs(){
	while(!Q.empty()){
		ll t=Q.front();
		Q.pop();
		if(t.x==T&&t.flg==0)return t.step;
		for(int i=h[t.x];i;i=edge[i].l){
			int y=edge[i].to;
			ll now=t;
			now.x=y;
			if(now.flg==0)now.step++;
			now.flg=(now.flg+1)%3;
			if(!mem[now.flg][now.x])Q.push({now}),mem[now.flg][now.x]=1;
		}
	}
	return -1;
}
int main(){
	memset(mem,0,sizeof(mem));
	int n,m,u,v;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)scanf("%d%d",&u,&v),Add(i,u,v);
	scanf("%d%d",&S,&T);
	Q.push({S,0,0});
	mem[0][S]=1;
	printf("%d\n",bfs());
	return 0;
}
### Prompt

Please create a solution in CPP to 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 M 100002
using namespace std;
struct lll{
	int l,to;
}edge[M];
struct ll{
	int x,flg,step;
};
queue<ll>Q;
int h[M],S,T;
bool mem[5][M];
void Add(int i,int st,int to){
	edge[i].l=h[st];
	edge[i].to=to;
	h[st]=i;
}
int bfs(){
	while(!Q.empty()){
		ll t=Q.front();
		Q.pop();
		if(t.x==T&&t.flg==0)return t.step;
		for(int i=h[t.x];i;i=edge[i].l){
			int y=edge[i].to;
			ll now=t;
			now.x=y;
			if(now.flg==0)now.step++;
			now.flg=(now.flg+1)%3;
			if(!mem[now.flg][now.x])Q.push({now}),mem[now.flg][now.x]=1;
		}
	}
	return -1;
}
int main(){
	memset(mem,0,sizeof(mem));
	int n,m,u,v;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)scanf("%d%d",&u,&v),Add(i,u,v);
	scanf("%d%d",&S,&T);
	Q.push({S,0,0});
	mem[0][S]=1;
	printf("%d\n",bfs());
	return 0;
}
```