#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int MOD = 1e9+7;

int main(){
	
	int n,m; cin>>n>>m;
	vector<vector<int>> e(n);
	for(int i=0; i<m; i++){
		int u,v; cin>>u>>v;
		e.at(--u).push_back(--v);
	}
	int s,t; cin>>s>>t;
	--s; --t;
	vector<vector<int>> d(n,vector<int>(3,-1));
	d.at(s).at(0) = 0;
	queue<pair<int,int>> q;
	q.push({s,0});
	while(!q.empty()){
		auto p = q.front();
		q.pop();
		for(auto i:e.at(p.first)){
			if(d.at(i).at((p.second+1)%3)==-1){
				d.at(i).at((p.second+1)%3) = d.at(p.first).at(p.second)+1;
				q.push({i,(p.second+1)%3});
			}
		}
	}
	if(d.at(t).at(0)==-1) d.at(t).at(0)=-3;
	cout << d.at(t).at(0)/3 << endl;
}


### Prompt

Your challenge is to write 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;
using ll = long long;
const int MOD = 1e9+7;

int main(){
	
	int n,m; cin>>n>>m;
	vector<vector<int>> e(n);
	for(int i=0; i<m; i++){
		int u,v; cin>>u>>v;
		e.at(--u).push_back(--v);
	}
	int s,t; cin>>s>>t;
	--s; --t;
	vector<vector<int>> d(n,vector<int>(3,-1));
	d.at(s).at(0) = 0;
	queue<pair<int,int>> q;
	q.push({s,0});
	while(!q.empty()){
		auto p = q.front();
		q.pop();
		for(auto i:e.at(p.first)){
			if(d.at(i).at((p.second+1)%3)==-1){
				d.at(i).at((p.second+1)%3) = d.at(p.first).at(p.second)+1;
				q.push({i,(p.second+1)%3});
			}
		}
	}
	if(d.at(t).at(0)==-1) d.at(t).at(0)=-3;
	cout << d.at(t).at(0)/3 << endl;
}


```