#include <bits/stdc++.h>
using namespace std;
#define int long long

int d[100005][3],s,t,n,m;
vector<int> g[100005];
queue<pair<int,int> >q;
signed main(){
	cin >> n >> m;
	for(int i = 1 ; i <= m ; i ++){
		int u,v;
		cin >> u >> v;
		g[u].push_back(v);
	}
	cin >> s >> t;
	q.push({s, 0});
	while(q.size()){
		int u = q.front().first, dis = q.front().second;
		q.pop();
		if(d[u][dis%3]) continue;
		d[u][dis%3] = 1;
		if(u == t && dis%3 == 0){
			cout << dis/3;
			return 0;
		}
		for(int v:g[u])
			if(!d[v][(dis+1)%3])
				q.push({v, dis + 1});
	}
	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 <bits/stdc++.h>
using namespace std;
#define int long long

int d[100005][3],s,t,n,m;
vector<int> g[100005];
queue<pair<int,int> >q;
signed main(){
	cin >> n >> m;
	for(int i = 1 ; i <= m ; i ++){
		int u,v;
		cin >> u >> v;
		g[u].push_back(v);
	}
	cin >> s >> t;
	q.push({s, 0});
	while(q.size()){
		int u = q.front().first, dis = q.front().second;
		q.pop();
		if(d[u][dis%3]) continue;
		d[u][dis%3] = 1;
		if(u == t && dis%3 == 0){
			cout << dis/3;
			return 0;
		}
		for(int v:g[u])
			if(!d[v][(dis+1)%3])
				q.push({v, dis + 1});
	}
	cout << -1;
}
```