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>>;
int main(){
  int N,M;
  cin>>N>>M;
  Graph G(N);
  for(int i=0;i<M;i++){
  int u,v; cin>>u>>v; u--;v--;
    G[u].push_back(v);
  }int s,t;cin>>s>>t; s--; t--;
  vector<vector<int>>dist(3,vector<int>(N,-1));
  dist[0][s]=0;
  vector<int>dist1(N,-1); dist1[s]=0;
  queue<pair<int,int>>que;pair<int,int>p(s,0);que.push(p);
  while(!que.empty()){
  pair<int,int>q=que.front();
    que.pop();int v=q.first;
    for(auto u:G[v]){
      int x=q.second+1;
    int i=x%3;
      if(dist[i][u]!=-1)
        continue;
      else{
      dist[i][u]=x;
        dist1[u]=x;
        pair<int,int>r(u,x);
        que.push(r);
      }
    }
  
  }if(dist[0][t]==-1)
    cout<<-1<<endl;
  else
    cout<<dist[0][t]/3<<endl;
  
  return 0;
}