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;
using ll = long long;
using pii = pair<int, int>;

const int MAX_N = 100010;
int N, M, s, t;
int d1[MAX_N], d2[MAX_N];
vector<int> G[MAX_N];
int cnt1[MAX_N], cnt2[MAX_N];

void bfs(int v, int *d) {
  queue<int> que;
  que.emplace(v);
  memset(d, -1, sizeof(int) * N);
  d[v] = 0;
  while (!que.empty()) {
    int curr = que.front(); que.pop();
    for (int next : G[curr]) {
      if (d[next] < 0) {
        d[next] = d[curr] + 1;
        que.emplace(next);
      }
    }
  }
}

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  cin >> N >> M >> s >> t;
  s--, t--;
  for (int i = 0; i < M; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    G[x].emplace_back(y);
    G[y].emplace_back(x);
  }
  bfs(s, d1);
  bfs(t, d2);
  for (int i = 0; i < N; i++) {
    if (d1[i] >= 0) cnt1[d1[i]]++;
    if (d2[i] >= 0) cnt2[d2[i]]++;
  }
  ll ans = 0;
  for (int i = 0; i < d1[t] - 1; i++) {
    ans += (ll)cnt1[i] * cnt2[d1[t] - 2 - i];
  }
  cout << ans << endl;

  return 0;
}
