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>
using namespace std;

#define fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define mp(a,b) make_pair(a,b)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<P, int> PPI;
typedef pair<ll, ll> PL;
typedef pair<P, ll> PPL;

#define INF INT_MAX/3

#define MAX_N 1000

int n,m,s,t;
vector<int> G[111111];

int sd[111111],td[111111];
bool vis[111111];

int main(){
	cin.sync_with_stdio(false);
	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);
	}
	int maxd;
	queue<P> que;
	que.push(P(s,0));
	memset(sd,0,sizeof(sd));
	vis[s]=true;
	while(!que.empty()){
		int nv=que.front().fi,dist=que.front().se; que.pop();
		sd[dist]++;
		if(nv==t){
			maxd=dist;
		}
		rep(i,G[nv].size()){
			int next=G[nv][i];
			if(vis[next])continue;
			vis[next]=true;
			que.push(P(next,dist+1));
		}
	}

	queue<P>().swap(que);
	memset(td,0,sizeof(td));
	memset(vis,0,sizeof(vis));
	que.push(P(t,0));
	vis[t]=true;
	while(!que.empty()){
		int nv=que.front().fi,dist=que.front().se; que.pop();
		td[dist]++;
		rep(i,G[nv].size()){
			int next=G[nv][i];
			if(vis[next])continue;
			vis[next]=true;
			que.push(P(next,dist+1));
		}
	}

	ll res=0;
	rep(i,maxd-1){
		res+=(ll)sd[i]*td[maxd-i-2];
	}
	cout<<res<<endl;
	return 0;
}