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 <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <map>
#define INF 1000000000
using namespace std;
struct po{int node,flg;};
typedef long long ll;
int n,m,s,g;
vector <int> G[100001];
map <int,int> mp[2];

ll bfs(){
  int D[2][100001]={};
  queue <po> Q;
  mp[0].clear(),mp[1].clear();
  mp[0][0]++,mp[1][0]++;
  Q.push(po{s,0}),Q.push(po{g,1});
  while(!Q.empty()){
    po t=Q.front();Q.pop();
    for(int i=0;i<G[t.node].size();i++) {
      if(D[t.flg][G[t.node][i]]!=0)continue;
      Q.push(po{G[t.node][i],t.flg});
      D[t.flg][G[t.node][i]] = D[t.flg][t.node]+1;
      mp[t.flg][D[t.flg][t.node]+1]++;
    }
  }
  mp[0][D[0][s]]--,mp[1][D[1][g]]--;
  int DG = D[0][g];
  ll sum=0;
  map<int,int>::iterator it=mp[0].begin();
  while(it!=mp[0].end()){
    if(mp[1].count(DG-2-it->first))sum += (ll)it->second * mp[1][DG - it->first-2];
    it++;
  }
  return sum;
}

int main() {
  cin >> n >> m >> s >>g;
  for(int i=0;i<m;i++){
    int a,b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  cout << bfs() <<endl;
  return 0;
}