#include<bits/stdc++.h>
#define N 100005
#define MO 1000000007
using namespace std;
int n,m,K,st,ed;
int dp[N][3];
vector<int>P[N];
struct node {
	int a,b,c;
};
void spfa(){
	memset(dp,0x3f,sizeof(dp));
	queue<node>Q;
	Q.push((node){st,0,0});
	dp[st][0]=0;
	while(!Q.empty()){
		node x=Q.front();
		Q.pop();
		if(dp[x.a][x.b]<x.c)continue;
		for(int i=0;i<P[x.a].size();i++){
			int a=P[x.a][i],b=(x.b+1)%3,c=x.c+(x.b+1)/3;
			if(dp[a][b]>c){
				dp[a][b]=c;
				Q.push((node){a,b,c});
			}
		}
	}
}
void solve(){
	scanf("%d %d",&n,&m);
	for(int a,b,i=1;i<=m;i++){
		scanf("%d %d",&a,&b);
		P[a].push_back(b);
	}
	scanf("%d %d",&st,&ed);
	spfa();
	if(dp[ed][0]==0x3f3f3f3f)dp[ed][0]=-1;
	printf("%d\n",dp[ed][0]);
}
int main(){
	solve();
	return 0;
}
### Prompt

Generate a cpp solution 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 N 100005
#define MO 1000000007
using namespace std;
int n,m,K,st,ed;
int dp[N][3];
vector<int>P[N];
struct node {
	int a,b,c;
};
void spfa(){
	memset(dp,0x3f,sizeof(dp));
	queue<node>Q;
	Q.push((node){st,0,0});
	dp[st][0]=0;
	while(!Q.empty()){
		node x=Q.front();
		Q.pop();
		if(dp[x.a][x.b]<x.c)continue;
		for(int i=0;i<P[x.a].size();i++){
			int a=P[x.a][i],b=(x.b+1)%3,c=x.c+(x.b+1)/3;
			if(dp[a][b]>c){
				dp[a][b]=c;
				Q.push((node){a,b,c});
			}
		}
	}
}
void solve(){
	scanf("%d %d",&n,&m);
	for(int a,b,i=1;i<=m;i++){
		scanf("%d %d",&a,&b);
		P[a].push_back(b);
	}
	scanf("%d %d",&st,&ed);
	spfa();
	if(dp[ed][0]==0x3f3f3f3f)dp[ed][0]=-1;
	printf("%d\n",dp[ed][0]);
}
int main(){
	solve();
	return 0;
}
```