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;
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n,m,u,v,s,t;
  cin >> n >> m;
  vector<vector<int>> adj(n),adj3(n);
  for(int i = 0; i < m; ++i){
    cin >> u >> v;
    adj[u-1].push_back(v-1);
  }
  cin >> s >> t;
  --s;
  --t;
  int ans = 0;
  vector<vector<bool>> vu(3,vector<bool>(n));
  vu[0][s] = true;
  queue<pair<int,int>> q;
  q.emplace(s,0);
  while(!q.empty()){
    auto p = q.front();
    q.pop();
    if(p.first == t and p.second%3 == 0){
      cout << p.second/3;
      return 0;
    }
    int nt = p.second+1,nt3 = nt%3;
    for(int i : adj[p.first]){
      if(!vu[nt3][i]){
        vu[nt3][i] = true;
        q.emplace(make_pair(i,nt));
      }
    }
  }
  cout << -1;
  return 0;
}