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,N) for(int i=(n);i<(int)N;i++)
#define p(s) cout<<(s)<<endl
#define ck(n,a,b) ((a)<=(n)&&(n)<(b))
#define F first
#define S second
typedef long long ll;
using namespace std;
const int inf=1e9;

ll d[100010];//dist
ll fromS[100010];
ll fromT[100010];
vector<ll> edge[100100];
int main() {
	ll N,M,s,t;
	cin>>N>>M>>s>>t;
	s--;t--;
	REP(i,0,M){
		ll x,y;
		cin>>x>>y;
		x--;y--;
		edge[x].push_back(y);
		edge[y].push_back(x);
	}
	queue<ll> q,qq;
	fromS[0]=1;fromT[0]=1;
	q.push(s);
	REP(i,0,N) d[i]=inf;
	d[s]=0;
	while(!q.empty()){
		int now=q.front();q.pop();
		for(int next:edge[now]){
			if(d[next]<inf) continue;
			q.push(next);
			d[next]=d[now]+1;
			fromS[d[next]]++;
		}
		if(d[t]<inf) break;
	}

	REP(i,0,N) d[i]=inf;
	qq.push(t);
	d[t]=0;
	while(!qq.empty()){
		int now=qq.front();qq.pop();
		for(auto next:edge[now]){
			if(d[next]<inf) continue;
			qq.push(next);
			d[next]=d[now]+1;
			fromT[d[next]]++;
		}
		if(d[s]<inf) break;
	}
	ll ans=0;
	REP(i,0,d[s]-1){
		ans+=fromS[i]*fromT[d[s]-2-i];
	}
	p(ans);
	return 0;
}