#include <bits/stdc++.h>
using namespace std;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n, m; cin >> n >> m;
  VV<> g(3 * n);
  auto _ = [&](int v, int i) -> int { return 3 * v + i % 3; };
  while (m--) {
    int u, v; cin >> u >> v, --u, --v;
    for (int i = 0; i < 3; ++i) {
      g[_(u, i)].push_back(_(v, i + 1));
    }
  }
  int s, t; cin >> s >> t, --s, --t;
  queue<int> que;
  V<> d(3 * n, -1);
  d[_(s, 0)] = 0;
  que.push(_(s, 0));
  while (!que.empty()) {
    int v = que.front(); que.pop();
    for (int w : g[v]) if (d[w] == -1) {
      d[w] = d[v] + 1;
      que.push(w);
    }
  }
  cout << (d[_(t, 0)] != -1 ? d[_(t, 0)] / 3 : -1) << '\n';
}
### Prompt

In Cpp, your task is to solve 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;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n, m; cin >> n >> m;
  VV<> g(3 * n);
  auto _ = [&](int v, int i) -> int { return 3 * v + i % 3; };
  while (m--) {
    int u, v; cin >> u >> v, --u, --v;
    for (int i = 0; i < 3; ++i) {
      g[_(u, i)].push_back(_(v, i + 1));
    }
  }
  int s, t; cin >> s >> t, --s, --t;
  queue<int> que;
  V<> d(3 * n, -1);
  d[_(s, 0)] = 0;
  que.push(_(s, 0));
  while (!que.empty()) {
    int v = que.front(); que.pop();
    for (int w : g[v]) if (d[w] == -1) {
      d[w] = d[v] + 1;
      que.push(w);
    }
  }
  cout << (d[_(t, 0)] != -1 ? d[_(t, 0)] / 3 : -1) << '\n';
}
```