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 <vector>
#include <queue>

using namespace std;

int ds[2][100001];
int ns[2][100001];
vector<vector<int>> G;
void bfs(int x, int d, int i){
  queue<int> q, dis;
  q.push(x);
  dis.push(0);

  while(!q.empty()){
	x = q.front(); q.pop();
	d = dis.front(); dis.pop();
	if(ds[i][x] >= 0) continue;
	ds[i][x] = d;
	ns[i][d]++;
	for(int to: G[x]){
	  q.push(to);
	  dis.push(d+1);
	}
  }
}

int main(){
  fill((int*)ds, (int*)ds+2*100001, -1);
  fill((int*)ns, (int*)ns+2*100001, 0);

  int N, M, S, T;
  cin >> N >> M >> S >> T;
  --S, --T;
  G.assign(N, vector<int>());
  for(int i=0;i<M;++i){
	int x, y; cin >> x >> y;
	--x, --y;
	G[x].push_back(y);
	G[y].push_back(x);
  }
  bfs(S, 0, 0);
  bfs(T, 0, 1);

  long long ans = 0, d = ds[0][T];
  for(int i=0;i<=d-2;++i)
	ans += (long long)ns[0][i] * ns[1][d-2-i];
  cout << ans << endl;
}