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<queue>
#include<algorithm>
#include<cstring>
#include<vector>
#include<map>
#include<set>

#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef long long ll;

ll n,m,s,t,x,y;
vector<ll> g[100100];
ll ds[100100], dt[100100];
vector<ll> nums[100100];
set<ll> numt[100100];
const ll INF = 1e12;

int main(){
  cin >> n >> m >> s >> t;
  rep(i,m){
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }

  queue<ll> q;
  q.push(s);
  rep(i,n+1)ds[i] = INF;
  ds[s] = 0;
  nums[0].push_back(s);

  while(q.size()){
    ll v = q.front(); q.pop();

    for(int i=0;i<g[v].size();i++){
      if(ds[g[v][i]] > ds[v] + 1){
	ds[g[v][i]] = ds[v] + 1;
	q.push(g[v][i]);
	nums[ds[g[v][i]]].push_back(g[v][i]);
      }
    }
  }

  q.push(t);
  rep(i,n+1)dt[i] = INF;
  dt[t] = 0;
  numt[0].insert(t);

  while(q.size()){
    ll v = q.front(); q.pop();
    for(int i=0;i<g[v].size();i++){
      if(dt[g[v][i]] > dt[v] + 1){
	dt[g[v][i]] = dt[v] + 1;
	q.push(g[v][i]);
	numt[dt[g[v][i]]].insert(g[v][i]);
      }
    }
  }


  ll res = 0;
  ll len = ds[t]-1;
  /*
  rep(i,len+1){
    cout << "---" << i << "---" << endl;
    rep(j,nums[i].size())cout << nums[i][j] << " ";cout << endl;
  }

  rep(i,len+1){
    cout << "---" << i << "---" << endl;
    for(set<ll>::iterator it=numt[i].begin();it!=numt[i].end();it++){
      cout << *it << " ";cout << endl;
    }
  }
  */
  for(int i=0;i<len;i++){
    res += (ll)nums[i].size()*numt[len-i-1].size();
    rep(j,nums[i].size()){
      rep(k,g[nums[i][j]].size()){
	int nxt = g[nums[i][j]][k];
	if(numt[len-i-1].find(nxt)!=numt[len-i-1].end()){
	  res--;
	}
      }
    }
  }
  cout << res << endl;
}