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<string>
#include<iomanip>
#include<cmath>
#include<vector>
#include<algorithm>
#include<queue>

using namespace std;

#define int long long
#define rep(i,n) for(int i = 0; i < (n); i++)
#define INF (long long)(1e18)
#define MOD (int)(1e9+7)

#define yn(f) ((f)?"Yes":"No")
#define YN(f) ((f)?"YES":"NO")

#define MAX 110000

vector<int> G[MAX];
int d[MAX], d2[MAX], con[MAX],con2[MAX];
int N, M, s, t, x, y;

void bfs(int ss, int d[]){
	fill(d,d+N+1,INF);
	queue<int> Q;
	
	d[ss] = 0;
	Q.push(ss);
	
	while(!Q.empty()){
		int f = Q.front(); Q.pop();//cout<<f<<endl;
		// if(f == t) break;
		for(int y : G[f]){
			if(d[y] == INF){
				d[y] = d[f]+1;
				Q.push(y);
			}
		}
	}
}

int solve(){
	int re = 0;
	bfs(s,d);bfs(t,d2);
	for(int i = 1; i <= N; i++){
		if(d[i] != INF)con[d[i]]++;
		if(d2[i] != INF)con2[d2[i]]++;
	}//cout<<" test test "<<endl;
	for(int i = 0; i <= d[t]-2; i++){
		// cout<<i<<" "<<con[i]<<" "<<con2[i]<<endl;
		re += con[i]*con2[d[t]-i-2];
	}
	return re;
}

signed main(){
	cout<<fixed<<setprecision(7);
	
	cin>>N>>M>>s>>t;
	
	

	rep(i,M){
		cin>>x>>y;
		G[x].push_back(y);
		G[y].push_back(x);
	}
	
	cout<<solve()<<endl;
	
	return 0;
}
