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

constexpr int maxn = 1e5 + 10, inf = 1e8;
int n, m;
vector<int> g[maxn];
int d[maxn][3];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, u, v; i < m; ++i) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
  }
  int s, t;
  scanf("%d%d", &s, &t);
  for (int i = 1; i <= n; ++i) {
    d[i][0] = d[i][1] = d[i][2] = inf;
  }
  d[s][0] = 0;
  queue<pair<int, int>> q;
  q.emplace(s, 0);
  while (q.size()) {
    int x = q.front().first;
    int t = q.front().second;
    q.pop();
    for (int y : g[x]) {
      int &tmp = d[y][(t + 1) % 3];
      if (tmp == inf) {
        tmp = d[x][t] + 1;
        q.emplace(y, (t + 1) % 3);
      }
    }
  }
  printf("%d\n",  (d[t][0] != inf) ? d[t][0] / 3 : -1);
  return 0;
}


### Prompt

Your task is to create 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;

constexpr int maxn = 1e5 + 10, inf = 1e8;
int n, m;
vector<int> g[maxn];
int d[maxn][3];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, u, v; i < m; ++i) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
  }
  int s, t;
  scanf("%d%d", &s, &t);
  for (int i = 1; i <= n; ++i) {
    d[i][0] = d[i][1] = d[i][2] = inf;
  }
  d[s][0] = 0;
  queue<pair<int, int>> q;
  q.emplace(s, 0);
  while (q.size()) {
    int x = q.front().first;
    int t = q.front().second;
    q.pop();
    for (int y : g[x]) {
      int &tmp = d[y][(t + 1) % 3];
      if (tmp == inf) {
        tmp = d[x][t] + 1;
        q.emplace(y, (t + 1) % 3);
      }
    }
  }
  printf("%d\n",  (d[t][0] != inf) ? d[t][0] / 3 : -1);
  return 0;
}


```