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
#include <iostream>
#include <queue>
#include <utility>

using namespace std;

typedef pair<int, int> P;

int n, m, s, t;
vector<int> edge[110000];
queue<P> q;
bool visited[110000];

int main()
{
  cin >> n >> m;
  
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    edge[u].push_back(v);
  }
  
  cin >> s >> t;
  
  q.push(make_pair(s, 0));
  visited[s] = true;
  
  while (q.size()) {
    for (int v: edge[q.front().first]) {
      for (int w: edge[v]) {
        for (int x: edge[w]) {
          if (visited[x]) continue;
          visited[x] = true;
          if (x == t) {
            cout << q.front().second+1 << endl;
            return 0;
          }
          q.push(make_pair(x, q.front().second+1));
        }
      }
    }
    q.pop();
  }
  
  cout << -1 << endl;
  return 0;
}