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;

#define MAX_V 100000
#define INF (1<<29)
typedef long long ll;

void bfs(int s,int *d,ll *cnt,vector<int> G[MAX_V]){
  queue<int> Q; Q.push(s);
  memset(cnt,0,sizeof(cnt));
  fill(d,d+MAX_V,INF);
  d[s] = 0;
  cnt[d[s]]++;
  while(!Q.empty()){
    int v = Q.front(); Q.pop();
    for(int i = 0 ; i < (int)G[v].size() ; i++){
      int to = G[v][i];
      if(d[v] + 1 < d[to]){
        d[to] = d[v] + 1;
        cnt[d[to]]++;
        Q.push(to);
      }
    }
  }
}

int main(){
  int V,E,S,T,x,y;
  int ds[MAX_V],dt[MAX_V];
  ll cs[MAX_V],ct[MAX_V];
  vector<int> G[MAX_V];
  cin >> V >> E >> S >> T;
  S--; T--;
  for(int i = 0 ; i < E ; i++){
    cin >> x >> y; x--; y--;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  
  bfs(S,ds,cs,G); 
  bfs(T,dt,ct,G);
  
  ll ans = 0LL;
  int min_dist = ds[T];
  for(int i = 0 ; i < V ; i++){
    int j = min_dist-2-i;
    ans += (j >= 0 ? cs[i]*ct[j] : 0);
  }
  cout << ans << endl;
  return 0;
}