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 ALL(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define mod 1000000007
using ll=long long;
const int INF=1000000000;
const ll LINF=1001002003004005006ll;
int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
// ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>bool chmin(T &a,const T &b){if(b<a){a=b;return true;}return false;}
 
struct IOSetup{
    IOSetup(){
        cin.tie(0);
        ios::sync_with_stdio(0);
        cout<<fixed<<setprecision(12);
    }
} iosetup;
 
template<typename T1,typename T2>
ostream &operator<<(ostream &os,const pair<T1,T2>&p){
    os<<p.first<<" "<<p.second;
    return os;
}
 
template<typename T>
ostream &operator<<(ostream &os,const vector<T>&v){
    for(int i=0;i<(int)v.size();i++) os<<v[i]<<(i+1==(int)v.size()?"":" ");
    return os;
}

template<typename T1,typename T2>
istream &operator>>(istream &is,pair<T1,T2>&p){
    is>>p.first>>p.second;
    return is;
}

template<typename T>
istream &operator>>(istream &is,vector<T>&v){
    for(T &x:v)is>>x;
    return is;
}


signed main(){
    int n,m,s,t;cin>>n>>m>>s>>t;s--,t--;
    vector<vector<int>> g(n);
    rep(i,m){
        int u,v;cin>>u>>v;u--,v--;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    vector<int> d(n,INF),rd(n,INF);
    {
        queue<int> que;
        que.push(s);d[s]=0;
        while(!que.empty()){
            int u=que.front();que.pop();
            for(auto to:g[u])if(chmin(d[to],d[u]+1))que.push(to);
        }
    }
    {
        queue<int> que;
        que.push(t);rd[t]=0;
        while(!que.empty()){
            int u=que.front();que.pop();
            for(auto to:g[u])if(chmin(rd[to],rd[u]+1))que.push(to);
        }
    }

    vector<ll> cnt(n,0);
    rep(i,n)if(rd[i]<n)cnt[rd[i]]++;
    ll res=0;
    rep(i,n){
        int op=d[t]-1-1-d[i];
        if(0<=op and op<n)res+=cnt[op];
    }
    cout<<res<<endl;
    return 0;
}

