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;
const int nmax=100005;
vector<vector<int>> d(nmax,vector<int>(3,-1));
vector<vector<int>> g(nmax);
int n,m,s,t;
long long bfs(){
  d[s][0]=0;
  queue<pair<int,int>> q;
  q.push(make_pair(s,0));
  while(!q.empty()){
    pair<int,int> p=q.front();
    q.pop();
    int u=p.first,v=p.second;
    for(auto x:g[u]){
      if(d[x][(v+1)%3]==-1){
        d[x][(v+1)%3]=d[u][v]+1;
        q.push(make_pair(x,(v+1)%3));
      }
    }
  }
  if(d[t][0]==-1){
    return -1;
  }
  else{
    return d[t][0]/3;
  }
}
int main(){
  cin >> n >> m;
  for(int i=0;i<m;i++){
    int a,b;cin >> a >> b;
    a--,b--;
    g[a].push_back(b);
  }
  cin >> s >> t;
  s--,t--;
  cout << bfs() << endl;
}