#include<iostream>
#include<vector>
#define DIM 300005
using namespace std;
int n, m, i, x, y, s, t, j, nod, vecin, p, u;
int c[DIM], d[DIM];
vector<int> v[DIM];
int main(){
    cin>> n >> m;
    for(i = 1; i <= m; i++){
        cin>> x >> y;
        for(j = 0; j < 3; j++){
            v[x * 3 + j].push_back(y * 3 + (j + 1) % 3);
        }
    }
    cin>> s >> t;
    d[s * 3] = 1;
    p = u = 1;
    c[1] = s * 3;
    while(p <= u){
        nod = c[p];
        p++;
        for(i = 0; i < v[nod].size(); i++){
            vecin = v[nod][i];
            if(d[vecin] == 0){
                d[vecin] = d[nod] + 1;
                c[++u] = vecin;
            }
        }
    }
    if(d[t * 3] == 0){
        cout<< -1;
    }
    else{
        cout<< d[t * 3] / 3;
    }
}

### Prompt

Your task is to create 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<iostream>
#include<vector>
#define DIM 300005
using namespace std;
int n, m, i, x, y, s, t, j, nod, vecin, p, u;
int c[DIM], d[DIM];
vector<int> v[DIM];
int main(){
    cin>> n >> m;
    for(i = 1; i <= m; i++){
        cin>> x >> y;
        for(j = 0; j < 3; j++){
            v[x * 3 + j].push_back(y * 3 + (j + 1) % 3);
        }
    }
    cin>> s >> t;
    d[s * 3] = 1;
    p = u = 1;
    c[1] = s * 3;
    while(p <= u){
        nod = c[p];
        p++;
        for(i = 0; i < v[nod].size(); i++){
            vecin = v[nod][i];
            if(d[vecin] == 0){
                d[vecin] = d[nod] + 1;
                c[++u] = vecin;
            }
        }
    }
    if(d[t * 3] == 0){
        cout<< -1;
    }
    else{
        cout<< d[t * 3] / 3;
    }
}

```