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 N, M, s, t;
vector< int > graph[100000];

void bfs(int s, vector< int >& v)
{
  v.assign(N, -1);
  queue< int > que;
  que.push(s);
  v[s] = 0;
  while(!que.empty()) {
    int p = que.front(); que.pop();
    for(int to : graph[p]) {
      if(v[to] == -1) {
        v[to] = v[p] + 1;
        que.push(to);
      }
    }
  }            
}
int main()
{
  scanf("%d %d %d %d", &N, &M, &s, &t);
  --s, --t;
  for(int i = 0; i < M; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    --x, --y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }

  vector< int > m1, m2;
  bfs(s, m1);
  bfs(t, m2);
  
  int sum1[100001] = {}, sum2[100001] = {};
  for(int k : m1) if(k >= 0) sum1[k]++;
  for(int k : m2) if(k >= 0) sum2[k]++;
  long long ret = 0;
  for(int i = 0; i < m1[t] - 1; i++) {
    ret += 1LL * sum1[i] * sum2[m1[t] - 2 - i];
  }
  printf("%lld\n", ret);
}