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>
#define int long long
using namespace std;

signed main(){
	int n,m,s,t,ds[100000],dt[100000],ans = 0,scnt[100000] = {},tcnt[100000] = {};
	vector<int> G[100000];
	queue<int> que;
	cin >> n >> m >> s >> t; s--;t--;
	for(int i = 0;i < m;i++){
		int a,b;
		cin >> a >> b; a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	for(int i = 0;i < n;i++){
		ds[i] = -1;
		dt[i] = -1;
	}
	ds[s] = 0;
	que.push(s);
	while(!que.empty()){
		int v = que.front();que.pop();
		for(int to : G[v]){
			if(ds[to] == -1){
				ds[to] = ds[v] + 1;
				que.push(to);
			}
		}
	}
	dt[t] = 0;
	que.push(t);
	while(!que.empty()){
		int v = que.front();que.pop();
		for(int to : G[v]){
			if(dt[to] == -1){
				dt[to] = dt[v] + 1;
				que.push(to);
			}
		}
	}
	for(int i = 0;i < n;i++){
		if(ds[i] >= 0) scnt[ds[i]]++;
		if(dt[i] >= 0) tcnt[dt[i]]++;
	}
	for(int i = 0;i < ds[t] - 1;i++){
		ans += scnt[i] * tcnt[ds[t] - i - 2];
	}
	cout << ans << endl;
	return 0;
}