#include <bits/stdc++.h>

using namespace std;

const int MAXN = 3e5 + 5;
const int INF = 1e9;

set <int> g[MAXN];
set <int> gt[MAXN];


int main(){

	int n, m; cin >> n >> m;


	int u, v;
	for(int i = 0; i < m; i++){
		cin >> u >> v;

		u--; v--;

		g[u].insert(v + n);
		g[u + n].insert(v + 2 * n);
		g[u + 2 * n].insert(v);
	}


	int s, t; cin >> s >> t;

	s--; t--;


	vector <int> d(n * 3, INF);

	queue <int> q;

	q.push(s);
	d[s] = 0;

	while(!q.empty()){
		int v = q.front(); q.pop();

		for(int to : g[v]){
			if(d[to] > d[v] + 1){
				d[to] = d[v] + 1;
				q.push(to);
			}
		}
	}

	cout << (d[t] == INF ? -1 : d[t] / 3) << endl;


	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 = 3e5 + 5;
const int INF = 1e9;

set <int> g[MAXN];
set <int> gt[MAXN];


int main(){

	int n, m; cin >> n >> m;


	int u, v;
	for(int i = 0; i < m; i++){
		cin >> u >> v;

		u--; v--;

		g[u].insert(v + n);
		g[u + n].insert(v + 2 * n);
		g[u + 2 * n].insert(v);
	}


	int s, t; cin >> s >> t;

	s--; t--;


	vector <int> d(n * 3, INF);

	queue <int> q;

	q.push(s);
	d[s] = 0;

	while(!q.empty()){
		int v = q.front(); q.pop();

		for(int to : g[v]){
			if(d[to] > d[v] + 1){
				d[to] = d[v] + 1;
				q.push(to);
			}
		}
	}

	cout << (d[t] == INF ? -1 : d[t] / 3) << endl;


	return 0;
}
```