#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+100;
bool vis[maxn][3]={0};
vector<int> G[maxn];
struct node{
	int id,dis;
};
queue<node>q; 
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	int a,b;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&a,&b);
		G[a].push_back(b);
	}
	int s,t;
	scanf("%d%d",&s,&t);
	q.push({s,0});
	vis[s][0]=1;
	while(!q.empty()){
		node u=q.front();
		q.pop();
		for(auto i:G[u.id]){
			if(vis[i][2])continue;
			vis[i][2]=1;
			for(auto j:G[i]){
				if(vis[j][1])continue;
				vis[j][1]=1;
				for(auto k:G[j]){
					if(vis[k][0])continue;
					vis[k][0]=1;
					if(k==t){
						printf("%d\n",u.dis+1);
						return 0;
					} 
					q.push({k,u.dis+1});
				}
			}
		}
	}
	printf("-1\n");
}


### 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>
using namespace std;
const int maxn = 1e5+100;
bool vis[maxn][3]={0};
vector<int> G[maxn];
struct node{
	int id,dis;
};
queue<node>q; 
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	int a,b;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&a,&b);
		G[a].push_back(b);
	}
	int s,t;
	scanf("%d%d",&s,&t);
	q.push({s,0});
	vis[s][0]=1;
	while(!q.empty()){
		node u=q.front();
		q.pop();
		for(auto i:G[u.id]){
			if(vis[i][2])continue;
			vis[i][2]=1;
			for(auto j:G[i]){
				if(vis[j][1])continue;
				vis[j][1]=1;
				for(auto k:G[j]){
					if(vis[k][0])continue;
					vis[k][0]=1;
					if(k==t){
						printf("%d\n",u.dis+1);
						return 0;
					} 
					q.push({k,u.dis+1});
				}
			}
		}
	}
	printf("-1\n");
}


```