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 rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

struct sta{
  int x;
  int pre;
  int ppre;
};

int main(){
  int n,m,s,t;
  cin >> n >> m >> s >> t;
  s--,t--;
  vector<vector<int>> e(n);
  int u,v;
  rep(i,m){
    cin >> u >> v;
    e[u-1].push_back(v-1);
    e[v-1].push_back(u-1);
  }
  int len;
  vector<LL> sd(n+1,0),td(n+1,0);
  vector<int> dist(n,-1);
  queue<pair<int,int>> q;
  q.push({s,0});
  dist[s]=0;
  while(!q.empty()){
    auto now = q.front();
    q.pop();
    for(int nx:e[now.first]){
      if(dist[nx]==-1){
        dist[nx]=now.second+1;
        q.push({nx,now.second+1});
      }else{ 
        if(dist[nx]==now.second+1){
        }
      }
    }
  }
  rep(i,n) if(dist[i]!=-1) sd[dist[i]]++;
  len=dist[t];
  dist=vector<int>(n,-1);
  q.push({t,0});
  dist[t]=0;
  while(!q.empty()){
    auto now = q.front();
    q.pop();
    for(int nx:e[now.first]){
      if(dist[nx]==-1){
        dist[nx]=now.second+1;
        q.push({nx,now.second+1});
      }else{ 
        if(dist[nx]==now.second+1){
        }
      }
    }
  }
  rep(i,n) if(dist[i]!=-1) td[dist[i]]++;
  LL ans=0;
  rep(i,len-1){
    if(len-2-i<0) break;
    ans+=sd[i]*td[len-2-i];
  }
  cout << ans << endl;
  return 0;
}

