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 maxn= 100010;
vector<int> adj[maxn * 3];
int vis[maxn * 3];

int main() {
  #ifdef LOCAL
  freopen("input.txt", "r", stdin);
  #endif
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  while (m--) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v + n);
    adj[u + n].push_back(v + 2 * n);
    adj[u + 2 * n].push_back(v);
  }

  int s, t;
  cin >> s >> t;

  queue<pair<int, int>> q;
  q.push(make_pair(s, 0));
  vis[s] = 1;
  while (!q.empty()) {
    auto top = q.front(); q.pop();
    if (top.first == t) {
      cout << top.second / 3;
      return 0;
    }
    for (auto v : adj[top.first]) {
      if (!vis[v]) {
        vis[v] = 1;
        q.push(make_pair(v, top.second + 1));
      }
    }
  }
  cout << -1;
  return 0;
}