#include <bits/stdc++.h>

using namespace std;

vector<int> AdjList[300050];

const int modemul = 100005;

int dist[300050];

int main(){
	int N, M;
	scanf("%d%d", &N, &M);
	
	for(int i = 0; i < M; i ++){
		int a, b;
		scanf("%d%d", &a, &b);
		AdjList[a].push_back(b+modemul);
		AdjList[a+modemul].push_back(b+modemul*2);
		AdjList[a+modemul*2].push_back(b);
	}
	
	int S, T;
	scanf("%d%d", &S, &T);
	
	for(int i = 0; i < 300049; i ++){
		dist[i] = -3;
	}
	queue<int> q;
	q.push(S);
	dist[S] = 0;
	
	while(!q.empty()){
		int u = q.front(); q.pop();
		for(int v: AdjList[u]){
			if(dist[v] == -3){
				dist[v] = dist[u]+1;
				q.push(v);
			}
		}
	}
	printf("%d", dist[T]/3);
}

### Prompt

Please formulate 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;

vector<int> AdjList[300050];

const int modemul = 100005;

int dist[300050];

int main(){
	int N, M;
	scanf("%d%d", &N, &M);
	
	for(int i = 0; i < M; i ++){
		int a, b;
		scanf("%d%d", &a, &b);
		AdjList[a].push_back(b+modemul);
		AdjList[a+modemul].push_back(b+modemul*2);
		AdjList[a+modemul*2].push_back(b);
	}
	
	int S, T;
	scanf("%d%d", &S, &T);
	
	for(int i = 0; i < 300049; i ++){
		dist[i] = -3;
	}
	queue<int> q;
	q.push(S);
	dist[S] = 0;
	
	while(!q.empty()){
		int u = q.front(); q.pop();
		for(int v: AdjList[u]){
			if(dist[v] == -3){
				dist[v] = dist[u]+1;
				q.push(v);
			}
		}
	}
	printf("%d", dist[T]/3);
}

```