#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

vector<vector<ll>> g(300010);
vector<ll> dis(300010);

int main(){
    ll n, m;
    cin >> n >> m;
    ll i, j, k;
    for(i = 0; i < m;i++){
        ll u, v;
        cin >> u >> v;
        g[u].push_back(n+v);
        g[u + n].push_back(2 * n + v);
        g[u + 2 * n].push_back(v);
    }

    ll s, t;
    cin >> s >> t;
    queue<ll> que;
    que.push(s);
    while(!que.empty()){
        ll x = que.front();
        que.pop();
        for(ll i:g[x]){
            if(dis[i]==0){
                dis[i] = dis[x] + 1;
                que.push(i);
            }
        }
    }
    cout << (dis[t] > 0 ? dis[t]/3 : -1) << endl;
    return 0;
}
### Prompt

Please provide a CPP coded solution 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;
typedef long long ll;

vector<vector<ll>> g(300010);
vector<ll> dis(300010);

int main(){
    ll n, m;
    cin >> n >> m;
    ll i, j, k;
    for(i = 0; i < m;i++){
        ll u, v;
        cin >> u >> v;
        g[u].push_back(n+v);
        g[u + n].push_back(2 * n + v);
        g[u + 2 * n].push_back(v);
    }

    ll s, t;
    cin >> s >> t;
    queue<ll> que;
    que.push(s);
    while(!que.empty()){
        ll x = que.front();
        que.pop();
        for(ll i:g[x]){
            if(dis[i]==0){
                dis[i] = dis[x] + 1;
                que.push(i);
            }
        }
    }
    cout << (dis[t] > 0 ? dis[t]/3 : -1) << endl;
    return 0;
}
```