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>
typedef long long LL;
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;
LL p = 1000000007ll;

int main(void)
{
  LL n,m;
  cin >> n >> m;
  vector<vector<LL>> nei(n);
  REP(i,m) {
    LL x,y;
    cin >> x >> y;
    x %= n;
    y %= n;
    nei[x].push_back(y);
  }
  LL s,t;
  cin >> s >> t;
  s %= n;
  t %= n;
  priority_queue<pair<LL,LL>> q;
  q.emplace(0,s);
  vector<LL> cost(n*3,-3ll);
  while(!q.empty()) {
    LL nc,pos;
    tie(nc,pos) = q.top(); q.pop();
    if(cost[pos]!=-3) continue;
    cost[pos] = -nc;
    for(auto x:nei[pos%n]) {
      q.emplace(nc-1,(-(nc-1))%3*n+x);
    }
  }
  cout << cost[t]/3 << endl;
  return 0;
}
