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

int main(void){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int N, M, u, v, S, T;
	cin >> N >> M;
	vector<int> G[N];
	vector<vector<int>> dist(N, std::vector<int> (3,-1));
	for(int i = 0; i < M; ++i){
		cin >> u >> v;
		u--; v--;
		G[u].push_back(v);
	}
	cin >> S >> T;
	S--; T--;
	queue<pair<int, int>> q;
	q.emplace(0,S);
	dist[S][0] = 0;
	while(!q.empty()){
		int v = q.front().second;
		int d = q.front().first;
		q.pop();
		for(auto x : G[v]){
			if(dist[x][(d+1)%3] == -1){
				dist[x][(d+1)%3] = d+1;
				q.emplace(d+1,x);
			}
		}
	}
	cout << (dist[T][0] != -1 ? dist[T][0]/3 : -1) << endl;
	return 0;
}

### Prompt

Create a solution in CPP for 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;

int main(void){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int N, M, u, v, S, T;
	cin >> N >> M;
	vector<int> G[N];
	vector<vector<int>> dist(N, std::vector<int> (3,-1));
	for(int i = 0; i < M; ++i){
		cin >> u >> v;
		u--; v--;
		G[u].push_back(v);
	}
	cin >> S >> T;
	S--; T--;
	queue<pair<int, int>> q;
	q.emplace(0,S);
	dist[S][0] = 0;
	while(!q.empty()){
		int v = q.front().second;
		int d = q.front().first;
		q.pop();
		for(auto x : G[v]){
			if(dist[x][(d+1)%3] == -1){
				dist[x][(d+1)%3] = d+1;
				q.emplace(d+1,x);
			}
		}
	}
	cout << (dist[T][0] != -1 ? dist[T][0]/3 : -1) << endl;
	return 0;
}

```