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 int ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pair<int, int> > vii;
#define rrep(i, m, n) for(int (i)=(m); (i)<(n);  (i)++)
#define erep(i, m, n) for(int (i)=(m); (i)<=(n); (i)++)
#define  rep(i, n)    for(int (i)=0; (i)<(n);  (i)++)
#define rrev(i, m, n) for(int (i)=(n)-1; (i)>=(m); (i)--)
#define erev(i, m, n) for(int (i)=(n); (i)>=(m); (i)--)
#define  rev(i, n)    for(int (i)=(n)-1; (i)>=0; (i)--)
#define vrep(i, c)    for(__typeof((c).begin())i=(c).begin(); i!=(c).end(); i++)
#define  ALL(v)       (v).begin(), (v).end()
#define pb            push_back
#define mp            make_pair
template<class T, class S> inline bool minup(T& m, const S x){ return m>(T)x ? (m=(T)x, true) : false; }
template<class T, class S> inline bool maxup(T& m, const S x){ return m<(T)x ? (m=(T)x, true) : false; }

static const int    INF = 1000000000;
static const ll     MOD = 1000000007LL;
static const double EPS = 1E-10;

int N, M;
int s, t;
int x, y;
vi G[111111];
int ds[111111];
int dt[111111];
ll ns[111111];
ll nt[111111];

int main(int argc, char *argv[])
{
  cin >> N >> M;
  cin >> s >> t;
  s -= 1;
  t -= 1;
  rep(i, M){
    cin >> x >> y;
    G[x - 1].pb(y - 1);
    G[y - 1].pb(x - 1);
  }

  rep(i, N) ds[i] = -1;
  ds[s] = 0;
  queue<int> ques;
  ques.push(s);
  while(!ques.empty()){
    int v = ques.front();  ques.pop();
    vrep(w, G[v]) if(ds[*w] < 0){
      ds[*w] = ds[v] + 1;
      ques.push(*w);
    }
  }

  rep(i, N) dt[i] = -1;
  dt[t] = 0;
  queue<int> quet;
  quet.push(t);
  while(!quet.empty()){
    int v = quet.front();  quet.pop();
    vrep(w, G[v]) if(dt[*w] < 0){
      dt[*w] = dt[v] + 1;
      quet.push(*w);
    }
  }

  ll res = 0LL;
  rep(i, N) ns[ds[i]] += 1;
  rep(i, N) nt[dt[i]] += 1;
  rep(i, ds[t] - 1) res += ns[i] * nt[ds[t] - i - 2];
  cout << res << endl;
  
  return 0;
}