#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define   ull unsigned long long 
const int MAX=2e5+9;
const long long mod=1e9+7;
using namespace std;
vector<int>adj[MAX];
int dp[MAX][4];
ll pre[MAX][101];
int main(){
	int n ,m;
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int u, v;
		cin>>u>>v;
		u--;
		v--;
		adj[u].push_back(v);
	}
	int s,t;
	cin>>s>>t;
	s--;
	t--;
	queue<pair<int , int> > q;
	q.push(make_pair(s, 0));
	while(!q.empty()){
		pair<int , int> p = q.front();
		q.pop();
		int now=p.first;
		int cnt=p.second;
		for(int i=0;i<adj[now].size();i++){
			int to=adj[now][i];
			if(dp[to][(cnt+1)%3])continue;
			dp[to][(cnt+1)%3]=dp[now][cnt]+1;
			q.push(make_pair(to , (cnt+1)%3));
		}
	}
	if(dp[t][0])cout<<dp[t][0]/3;
	else
		cout<<-1;


}
### Prompt

In Cpp, your task is to solve 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 <iostream>
#include<bits/stdc++.h>
#define ll long long
#define   ull unsigned long long 
const int MAX=2e5+9;
const long long mod=1e9+7;
using namespace std;
vector<int>adj[MAX];
int dp[MAX][4];
ll pre[MAX][101];
int main(){
	int n ,m;
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int u, v;
		cin>>u>>v;
		u--;
		v--;
		adj[u].push_back(v);
	}
	int s,t;
	cin>>s>>t;
	s--;
	t--;
	queue<pair<int , int> > q;
	q.push(make_pair(s, 0));
	while(!q.empty()){
		pair<int , int> p = q.front();
		q.pop();
		int now=p.first;
		int cnt=p.second;
		for(int i=0;i<adj[now].size();i++){
			int to=adj[now][i];
			if(dp[to][(cnt+1)%3])continue;
			dp[to][(cnt+1)%3]=dp[now][cnt]+1;
			q.push(make_pair(to , (cnt+1)%3));
		}
	}
	if(dp[t][0])cout<<dp[t][0]/3;
	else
		cout<<-1;


}
```