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;
#define rep(i, N) for (int i = 0; i < (int)N; i++)

struct cand {
  int u, n;
};

vector<vector<int>> dist;
vector<vector<int>> adj;
queue<cand> q;

void kkp(int n, int i, int u) {
  if(dist[u][i] != -1) return;

  dist[u][i] = n;
  if(i == 0) {
    q.push(cand{u, n});
    return;
  }

  for(auto v: adj[u]) kkp(n, i-1, v);
}

int main () {
  int N, M;
  cin >> N >> M;
  adj = vector<vector<int>>(N+1);

  rep(i,M) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
  }

  int S, T;
  cin >> S >> T;

  dist = vector<vector<int>>(N+1, vector<int>(4, -1));
  q.push(cand{S, 0});

  while(!q.empty()) {
    cand c = q.front(); q.pop();
    kkp(c.n + 1, 3, c.u);
  }

  cout << dist[T][0] << endl;

  return 0;
}
