Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2
#include <bits/stdc++.h>
using namespace std;
int INF = 10000000;
int main(){
  int N, M, s, t;
  cin >> N >> M >> s >> t;
  s--;
  t--;
  vector<vector<int>> E(N);
  for (int i = 0; i < M; i++){
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    E[x].push_back(y);
    E[y].push_back(x);
  }
  vector<int> d1(N, INF);
  d1[s] = 0;
  queue<int> Q1;
  Q1.push(s);
  while (!Q1.empty()){
    int v = Q1.front();
    Q1.pop();
    for (int w : E[v]){
      if (d1[w] == INF){
        d1[w] = d1[v] + 1;
        Q1.push(w);
      }
    }
  }
  vector<int> d2(N, INF);
  d2[t] = 0;
  queue<int> Q2;
  Q2.push(t);
  while (!Q2.empty()){
    int v = Q2.front();
    Q2.pop();
    for (int w : E[v]){
      if (d2[w] == INF){
        d2[w] = d2[v] + 1;
        Q2.push(w);
      }
    }
  }
  int d = d1[t];
  vector<int> cnt1(N, 0);
  for (int i = 0; i < N; i++){
    if (d1[i] != INF){
      cnt1[d1[i]]++;
    }
  }
  vector<int> cnt2(N, 0);
  for (int i = 0; i < N; i++){
    if (d2[i] != INF){
      cnt2[d2[i]]++;
    }
  }
  long long ans = 0;
  for (int i = 0; i <= d - 2; i++){
    ans += (long long) cnt1[i] * cnt2[d - 2 - i];
  }
  cout << ans << endl;
}
