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;

typedef long long ll;
typedef vector<ll> v1;
typedef unordered_map<ll,unordered_map<ll,ll>> graph;

ll n,m;
graph g;
queue<v1> q;
ll s,t;
typedef vector<v1> v2;
v2 visit;

int main(){
  cin >> n >> m;
  for(ll i = 0;i < m;i++){
    ll u,v;
    cin >> u >> v;
    g[u][v] = 1;
  }
  cin >> s >> t;
  visit = v2(3,v1(n+1));
  q.push(v1({s, 0, 0}));
  visit[0][s] = 1;
  while(!q.empty()){
    ll x = q.front()[0];
    ll c = q.front()[1];
    ll d = q.front()[2];
    q.pop();
    if(x == t && c == 0){
      cout << d/3 << endl;
      return 0;
    }
    for(auto y : g[x]){
      if(visit[(c+1)%3][y.first]) continue;
      visit[(c+1)%3][y.first] = 1;
      q.push(v1({y.first, (c+1)%3, d+1}));
    }
  }
  cout << -1 << endl;
  return 0;


}
