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 <bits/stdc++.h>
using namespace std;
 
using Graph = vector< vector< int > > ;
 
const int64_t smax = 1001001 * 3; 
vector<int64_t> seen(100002*3, smax);

 
int main() {
  int N,M,S,T;
  cin >> N >> M;
 
  Graph G((N+1)*3);
  for (int i=0;i<M;++i ) {
    int a,b;
    cin >> a >> b;
//    --a; --b;
    G[3*a].push_back(3*b+1);
    G[3*a+1].push_back(3*b+2);
    G[3*a+2].push_back(3*b);
  }
  cin >> S >> T;
//  --S; --T;
  queue<int> q;
  q.push(S*3);
  seen[S*3] = 0;
  while( !q.empty() ) {
    int v = q.front();
    q.pop();
    for( auto next_v : G[v] ) {
      if (seen[next_v] != smax) continue;
      seen[next_v] = seen[v] + 1;
      q.push(next_v);
    }
  }
  if (seen[T*3] != smax) cout << seen[T*3]/3 << endl;
  else cout << "-1" << endl;
}
