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_N 100005
#define INF (1<<26)

int N,M,a,b,si,ti;
vector<int> G[MAX_N];
int sd[MAX_N],td[MAX_N];
int sc[MAX_N],tc[MAX_N];

void calc(int si,int d[MAX_N],int c[MAX_N]){
  for(int i=0;i<MAX_N;i++)d[i]=INF,c[i]=0;

  queue<int> Q;
  d[si]=0;
  c[0]++;
  Q.push(si);
  while(!Q.empty()){
    int pos=Q.front();Q.pop();
    for(int i=0;i<(int)G[pos].size();i++){
      int to=G[pos][i];
      if(d[pos]+1<d[to]){
	d[to]=d[pos]+1;
	c[d[to]]++;
	Q.push(to);
      }
    }
  }
}

int main(){
  scanf("%d %d %d %d",&N,&M,&si,&ti);
  for(int i=0;i<M;i++){
    scanf("%d %d",&a,&b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  calc(si,sd,sc);
  calc(ti,td,tc);
 
  int total=sd[ti];
  long long ans=0,A,B;
  for(int i=0;i<=N;i++){
    if(total-2-i<0)continue;
    A=sc[i];
    B=tc[total-2-i];
    ans+=A*B;
  }
  cout<<ans<<endl;
  return 0;
}