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

#define LL long long
const int N = 1e5 + 5;
int n, m, x, y, s, t, vis[3][N];
queue<pair<int,int>> q;
vector<int> g[N];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    cin >> n >> m;
    while(m--) {
        cin >> x >> y;
        g[x].push_back(y);
    }    

    cin >> s >> t;
    q.push({s, 0});
    vis[0][s] = 1;

    while(!q.empty()) {
        pair<int,int> u = q.front();
        q.pop();
        if(u.first == t && u.second % 3 == 0) return cout << u.second / 3, 0;

        int d = u.second + 1;
        for(int v : g[u.first]) if(vis[d % 3][v] == 0) vis[d % 3][v]++, q.push({v, d});
    }
    cout << -1;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 LL long long
const int N = 1e5 + 5;
int n, m, x, y, s, t, vis[3][N];
queue<pair<int,int>> q;
vector<int> g[N];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    cin >> n >> m;
    while(m--) {
        cin >> x >> y;
        g[x].push_back(y);
    }    

    cin >> s >> t;
    q.push({s, 0});
    vis[0][s] = 1;

    while(!q.empty()) {
        pair<int,int> u = q.front();
        q.pop();
        if(u.first == t && u.second % 3 == 0) return cout << u.second / 3, 0;

        int d = u.second + 1;
        for(int v : g[u.first]) if(vis[d % 3][v] == 0) vis[d % 3][v]++, q.push({v, d});
    }
    cout << -1;
}
```