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 int long long
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define all(v) (v).begin(),(v).end()
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define pb push_back
#define fi first
#define se second
template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N,M,S,T;
vector<int>G[100000];

vint bfs(int s){
    vint dist(N,1001001001);
    dist[s]=0;
    queue<int>que;
    que.push(s);
    while(que.size()){
        int v=que.front();
        que.pop();
        for(auto u:G[v]){
            if(dist[u]!=1001001001)continue;
            dist[u]=dist[v]+1;
            que.push(u);
        }
    }
    return dist;
}

signed main(){
    cin>>N>>M>>S>>T;
    S--;T--;
    rep(i,M){
        int a,b;
        cin>>a>>b;
        a--;b--;
        G[a].pb(b);G[b].pb(a);
    }

    vint distS=bfs(S);
    vint distT=bfs(T);

    int D=distS[T]-1;

    vint cntS(N),cntT(N);
    rep(i,N){
        if(distS[i]<N)cntS[distS[i]]++;
        if(distT[i]<N)cntT[distT[i]]++;
    }

    int ans=0;
    for(int i=0;i<D;i++){
        ans+=cntS[i]*cntT[D-i-1];
    }
    cout<<ans<<endl;
    return 0;
}