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<iostream>
#include<vector>
#include<map>
#include<queue>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
#define MAX 100005
int main(){
  int n,m,s,t,di;
  cin >> n >> m >> s >> t;
  int i,j,k;
  vector<int> v[MAX];
  ll d[MAX]={},d2[MAX]={};
  for(i=0;i<m;i++){
    cin >> j >> k;
    v[j].push_back(k);
    v[k].push_back(j);
  }

  bool used[MAX]={},used2[MAX]={};
  queue<P> q;
  P p,b;
  p.first=s;
  p.second=0;
  q.push(p);
  while(!q.empty()){
    p=q.front();q.pop();
    if(used[p.first]) continue;
    //cout << p.first << ":" << p.second << endl;
    if(p.first==t) di=p.second;
    used[p.first]=true;
    d[p.second]++;
    b.second=p.second+1;
    for(i=0;i<v[p.first].size();i++){
      b.first=v[p.first][i];
      q.push(b);
    }
  }
  p.first=t;
  p.second=0;
  q.push(p);
  while(!q.empty()){
    p=q.front();q.pop();
    if(used2[p.first]) continue;
    //cout << p.first << ":" << p.second << endl;
    used2[p.first]=true;
    d2[p.second]++;
    b.second=p.second+1;
    for(i=0;i<v[p.first].size();i++){
      b.first=v[p.first][i];
      q.push(b);
    }
  }
  
  //cout << di << endl;
  ll o=0;
  
  for(i=0;i<di-1;i++){
    o+=d[i]*d2[di-2-i];
  }
  
  cout << o << endl;
  return 0;
}