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

int main() {
    int n, m; cin >> n >> m;
    vector<int> e[3*n];
    for (int i = 0; i < m; i++) {
        int u, v; cin >> u >> v;
        u--; v--;
        e[u].emplace_back(v+n);
        e[u+n].emplace_back(v+n*2);
        e[u+n*2].emplace_back(v);
    }
    int s, t; cin >> s >> t; s--; t--;
    vector<int> d(3*n, 1e9);
    d[s] = 0;
    queue<int> q; q.push(s);
    while (!q.empty()) {
        int curr = q.front(); q.pop();
        for (auto &to: e[curr]) {
            if (d[to] > d[curr] + 1) {
                d[to] = d[curr] + 1;
                q.push(to);
            }
        }
    }
    if (d[t] != 1e9) cout << d[t] / 3 << endl;
    else cout << -1 << endl;
    return 0;
}
### Prompt

Please create a solution in cpp 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;

int main() {
    int n, m; cin >> n >> m;
    vector<int> e[3*n];
    for (int i = 0; i < m; i++) {
        int u, v; cin >> u >> v;
        u--; v--;
        e[u].emplace_back(v+n);
        e[u+n].emplace_back(v+n*2);
        e[u+n*2].emplace_back(v);
    }
    int s, t; cin >> s >> t; s--; t--;
    vector<int> d(3*n, 1e9);
    d[s] = 0;
    queue<int> q; q.push(s);
    while (!q.empty()) {
        int curr = q.front(); q.pop();
        for (auto &to: e[curr]) {
            if (d[to] > d[curr] + 1) {
                d[to] = d[curr] + 1;
                q.push(to);
            }
        }
    }
    if (d[t] != 1e9) cout << d[t] / 3 << endl;
    else cout << -1 << endl;
    return 0;
}
```