#include<bits/stdc++.h>
using namespace std;
#define M 100005
vector<int>d[M];
int dis[3][M];
struct hh{
	int p,md,w;
	bool operator<(hh a)const{
		return w>a.w;
	}
};
priority_queue<hh>q;
int main(){
	memset(dis,127,sizeof(dis));
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1,a,b;i<=m;++i){
		scanf("%d%d",&a,&b);
		d[a].push_back(b);
	}
	int s,t;
	scanf("%d%d",&s,&t);
	dis[0][s]=0;
	q.push((hh){s,0,0});
	while(!q.empty()){
		hh st=q.top();q.pop();
		if(dis[st.md][st.p]<st.w)continue;
		for(int i=0;i<(int)d[st.p].size();++i){
			int v=d[st.p][i];
			if((st.w+(st.md==2))<dis[(st.md+1)%3][v]){
				dis[(st.md+1)%3][v]=st.w+(st.md==2);
				q.push((hh){v,(st.md+1)%3,st.w+(st.md==2)});
			}
		}
	}
	if(dis[0][t]>1e9)dis[0][t]=-1;
	printf("%d",dis[0][t]);
}
### 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;
#define M 100005
vector<int>d[M];
int dis[3][M];
struct hh{
	int p,md,w;
	bool operator<(hh a)const{
		return w>a.w;
	}
};
priority_queue<hh>q;
int main(){
	memset(dis,127,sizeof(dis));
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1,a,b;i<=m;++i){
		scanf("%d%d",&a,&b);
		d[a].push_back(b);
	}
	int s,t;
	scanf("%d%d",&s,&t);
	dis[0][s]=0;
	q.push((hh){s,0,0});
	while(!q.empty()){
		hh st=q.top();q.pop();
		if(dis[st.md][st.p]<st.w)continue;
		for(int i=0;i<(int)d[st.p].size();++i){
			int v=d[st.p][i];
			if((st.w+(st.md==2))<dis[(st.md+1)%3][v]){
				dis[(st.md+1)%3][v]=st.w+(st.md==2);
				q.push((hh){v,(st.md+1)%3,st.w+(st.md==2)});
			}
		}
	}
	if(dis[0][t]>1e9)dis[0][t]=-1;
	printf("%d",dis[0][t]);
}
```