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;
const int INF = 1e9;
using PP = pair<int, int>;
int n, m, s, t;
vector<int> G[300000];

int bfs() {
  vector<int> dist(3 * n, INF);
  queue<PP> Q;
  Q.push(PP(0, s));
  while (!Q.empty()) {
    PP p = Q.front(); Q.pop();
    int d = p.first;
    int from = p.second;
    if (dist[from] != INF) continue;
    dist[from] = d;
    for (int to : G[from]) {
      Q.push(PP(d + 1, to));
    }
  }
  return dist[t];
}

int main() {
  cin >> n >> m;
  for (int j = 0; j < m; ++j) {
    int a, b;
    cin >> a >> b;
    --a; --b;
    for (int k = 0; k < 3; ++k) {
      G[a + k * n].push_back(b + (k + 1) % 3 * n);
    }
  }
  cin >> s >> t;
  --s; --t;
  int ans = bfs();
  cout << (ans == INF ? -1 : ans / 3) << endl;
}