#include <bits/stdc++.h>

using namespace std;

const int N = 100100;

vector<int> g[N];
int d[N][3];

int main() {
   ios_base::sync_with_stdio(false);
   int n, m;
   cin >> n >> m;
   for (int i = 1; i <= m; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
   }
   int s, t;
   cin >> s >> t;
   memset(d, -1, sizeof d);
   queue<pair<int, int>> q;
   d[s][0] = 0;
   q.emplace(s, 0);
   while (!q.empty()) {
      int u, r;
      tie(u, r) = q.front();
      q.pop();
      for (int v : g[u]) {
         int nr = (r + 1) % 3;
         if (d[v][nr] == -1) {
            d[v][nr] = d[u][r] + 1;
            q.emplace(v, nr);
         }
      }
   }
   if (d[t][0] == -1) {
      cout << -1;
   } else {
      cout << d[t][0] / 3 << "\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;

const int N = 100100;

vector<int> g[N];
int d[N][3];

int main() {
   ios_base::sync_with_stdio(false);
   int n, m;
   cin >> n >> m;
   for (int i = 1; i <= m; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
   }
   int s, t;
   cin >> s >> t;
   memset(d, -1, sizeof d);
   queue<pair<int, int>> q;
   d[s][0] = 0;
   q.emplace(s, 0);
   while (!q.empty()) {
      int u, r;
      tie(u, r) = q.front();
      q.pop();
      for (int v : g[u]) {
         int nr = (r + 1) % 3;
         if (d[v][nr] == -1) {
            d[v][nr] = d[u][r] + 1;
            q.emplace(v, nr);
         }
      }
   }
   if (d[t][0] == -1) {
      cout << -1;
   } else {
      cout << d[t][0] / 3 << "\n";
   }
}

```