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>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
int inf=1e8;
int cs[100000],ct[100000];
vector<int> G[100000];
int ds[100000],dt[100000];
int N;
void bfs(int s,int* d){
	queue<int> que;
	que.push(s);
	rep(i,N) d[i]=inf;
	d[s]=0;
	while(!que.empty()){
		int v=que.front();
		que.pop();
		for(int u:G[v]){
			if(d[u]==inf){
				que.push(u);
				d[u]=d[v]+1;
			}
		}
	}
}
int main(){
	int M,s,t;
	cin>>N>>M>>s>>t;
	s--,t--;
	rep(i,M){
		int x,y;
		cin>>x>>y;
		x--,y--;
		G[x].pb(y),G[y].pb(x);
	}
	bfs(s,ds);
	bfs(t,dt);
	int D=ds[t];
	rep(i,N) if(ds[i]!=inf) cs[ds[i]]++;
	rep(i,N) if(dt[i]!=inf) ct[dt[i]]++;
	long long ans=0;
	rep(i,D-1) ans+=(long long)cs[i]*ct[D-2-i];
	cout<<ans<<endl;
}