#include <bits/stdc++.h>

using namespace std;

int main(int argc, const char *argv[]) {
  int n, m;
  cin >> n >> m;

  vector<vector<int>> graph(n);
  for (int i = 0; i < m; ++i) {
    int src, dest;
    cin >> src >> dest;
    src--;
    dest--;
    graph[src].push_back(dest);
  }

  int s, t;
  cin >> s >> t;
  s--;
  t--;

  int ans = -1;
  vector<vector<int>> dists(n, vector<int>(3, -1));
  queue<pair<int, int>> q;
  q.push({s, 0});
  while (!q.empty()) {
    int node_idx = q.front().first, dist = q.front().second;
    q.pop();

    if (dists[node_idx][dist % 3] != -1) {
      continue;
    }

    dists[node_idx][dist % 3] = dist;
    for (auto &neighbor : graph[node_idx]) {
      q.push({neighbor, dist + 1});
    }
  }

  cout << (dists[t][0] < 0 ? dists[t][0] : dists[t][0] / 3) << '\n';

  return 0;
}

### Prompt

Please formulate a CPP solution 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 argc, const char *argv[]) {
  int n, m;
  cin >> n >> m;

  vector<vector<int>> graph(n);
  for (int i = 0; i < m; ++i) {
    int src, dest;
    cin >> src >> dest;
    src--;
    dest--;
    graph[src].push_back(dest);
  }

  int s, t;
  cin >> s >> t;
  s--;
  t--;

  int ans = -1;
  vector<vector<int>> dists(n, vector<int>(3, -1));
  queue<pair<int, int>> q;
  q.push({s, 0});
  while (!q.empty()) {
    int node_idx = q.front().first, dist = q.front().second;
    q.pop();

    if (dists[node_idx][dist % 3] != -1) {
      continue;
    }

    dists[node_idx][dist % 3] = dist;
    for (auto &neighbor : graph[node_idx]) {
      q.push({neighbor, dist + 1});
    }
  }

  cout << (dists[t][0] < 0 ? dists[t][0] : dists[t][0] / 3) << '\n';

  return 0;
}

```