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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define USE_LLONG_AS_INT
#ifdef USE_LLONG_AS_INT
#define int long long
#define inf (1ll<<60)
#else
#define inf (1<<30)
#endif
#define rep(i,n) for(int i=0;i<n;i++)
#define Rep(i,a,b) for(int i=a;i<b;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define rev(i,n) for(int i=(n)-1;i>=0;i--)
#define vi vector<int>
#define vvi vector<vi>
#define pb push_back
#define pi pair<int,int>
#define vp vector<pair<int,int>>
#define mp make_pair
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
#define MEMSET(a) memset(a,0,sizeof(a))
#define Yes(f) cout<<(f?"Yes":"No")<<endl
#define yes(f) cout<<(f?"yes":"no")<<endl
#define YES(f) cout<<(f?"YES":"NO")<<endl
#define SORT(v) sort(all(v))
#define RSORT(v) sort(all(v), greater<int>())

using namespace std;

const int mod=1e9+7;
const string sp=" ";

void run();

void init() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout<<fixed<<setprecision(12);
}

signed main(){
    init();
    run();
    return 0;
}

vector<vp> g;
int n,m;

vi dijkstra(int s){
    priority_queue<pi,vp,greater<pi>> q;
    vi d(n,inf);
    d[s]=0;
    q.push({0,s});
    while(!q.empty()){
        pi p=q.top();q.pop();
        int v=p.se;
        if(d[v]<p.fi)continue;
        rep(i,g[v].size()){
            pi e=g[v][i];
            if(d[e.fi]>d[v]+e.se){
                d[e.fi]=d[v]+e.se;
                q.push({d[e.fi],e.fi});
            }
        }
    }
    return d;
}

int solve(int s,int t){
    vi d1=dijkstra(s),d2=dijkstra(t);
    int l=d1[t];
    vi a1(l+1),a2(l+1);
    rep(i,n){
        if(d1[i]<=l)a1[d1[i]]++;
        if(d2[i]<=l)a2[d2[i]]++;
    }
    int ret=0;
    REP(i,0,l-2){
        ret+=a1[i]*a2[l-i-2];
    }
    return ret;
}

void run(){
    int s,t;
    cin>>n>>m>>s>>t;
    s--,t--;
    g=vector<vp>(n);
    rep(i,m){
        int x,y;
        cin>>x>>y;
        x--,y--;
        g[x].pb({y,1});
        g[y].pb({x,1});
    }
    cout<<solve(s,t)<<endl;
}
