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;
typedef long long ll;
ll c[2][100010],d[100010],INF=1<<29;
vector<ll> v[100010];
ll n,m,s,t;

void bfs(ll s,ll k) {
  fill(d,d+n+1,INF);
  d[s]=0;
  c[k][0]++;
  queue<ll> que;
  que.push(s);
  while(!que.empty()) {
    ll x=que.front();que.pop();
    for(int i=0; i<v[x].size(); i++) {
      ll y=v[x][i];
      if(d[y]<=d[x]+1) continue;
      d[y]=d[x]+1;
      c[k][d[y]]++;
      que.push(y);
    }
  }
}
  

int main() {
  cin >> n >> m >> s >> t;
  for(ll i=0; i<m; i++) {
    ll x,y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  memset(c,0,sizeof(c));
  bfs(s,0);
  bfs(t,1);
  ll z=d[s],ans=0;
  for(ll i=0; i<z-1; i++) ans+=c[0][i]*c[1][z-i-2];
  cout << ans << endl;
  return 0;
}