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 pair<ll,ll> pll;

int main() {
  ll n,m;
  cin >> n >> m;
  vector<set<ll>> v(n);
  for(ll i=0;i<m;i++) {
    ll a,b;
    cin >> a >> b;
    a--;
    b--;
    v[a].insert(b);
  }
  ll s,t;
  cin >> s >> t;
  s--;
  t--;
  vector<vector<ll>> dis(n,vector<ll> (3,-1));
  queue<pll> q;
  q.push(pll(s,0));
  dis[s][0]=0;
  while(q.size()) {
    ll f=q.front().first;
    ll d=q.front().second;
    q.pop();
    for(ll t:v[f]) {
      if(dis[t][(d+1)%3]<0) {
        dis[t][(d+1)%3]=d+1;
        q.push(pll(t,d+1));
      }
    }
  }
  if(dis[t][0]<0) {
    cout << -1 << endl;
  }
  else {
    cout << dis[t][0]/3 << endl;
  }
}