#include<bits/stdc++.h>
using namespace std;
const int M=100001;
int n,m,a,b,mi[M],s,t;
vector<int>d[M];
struct node{
	int x,step,cnt;
};
queue<node>Q;
bool mark[M][5];
int main(){
	scanf("%d%d",&n,&m);
	memset(mi,32,sizeof(mi));
	for(int i=1;i<=m;i++){
		scanf("%d%d",&a,&b);
		d[a].push_back(b);
	}
	scanf("%d%d",&s,&t);
	Q.push((node)<%s,0,0%>);
	mark[s][0]=1;
	while(!Q.empty()){
		node a=Q.front();
		Q.pop();
		if(a.step==0&&a.x==t){
			printf("%d\n",a.cnt/3);
			return 0;
		}
		int nst=(a.step+1)%3,ncnt=a.cnt+1;
		for(int i=0;i<(int)d[a.x].size();i++){
			int y=d[a.x][i];
			if(!mark[y][nst]){
				Q.push((node)<%y,nst,ncnt%>);
				mark[y][nst]=1;
			}
		}
	}
	puts("-1");
}
### 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>
using namespace std;
const int M=100001;
int n,m,a,b,mi[M],s,t;
vector<int>d[M];
struct node{
	int x,step,cnt;
};
queue<node>Q;
bool mark[M][5];
int main(){
	scanf("%d%d",&n,&m);
	memset(mi,32,sizeof(mi));
	for(int i=1;i<=m;i++){
		scanf("%d%d",&a,&b);
		d[a].push_back(b);
	}
	scanf("%d%d",&s,&t);
	Q.push((node)<%s,0,0%>);
	mark[s][0]=1;
	while(!Q.empty()){
		node a=Q.front();
		Q.pop();
		if(a.step==0&&a.x==t){
			printf("%d\n",a.cnt/3);
			return 0;
		}
		int nst=(a.step+1)%3,ncnt=a.cnt+1;
		for(int i=0;i<(int)d[a.x].size();i++){
			int y=d[a.x][i];
			if(!mark[y][nst]){
				Q.push((node)<%y,nst,ncnt%>);
				mark[y][nst]=1;
			}
		}
	}
	puts("-1");
}
```