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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

int main(){
    int n,m,s,t;
    cin>>n>>m>>s>>t;
    --s;--t;
    vector<int> v[n];
    rep(i,m){
        int x,y;
        cin>>x>>y;
        --x;--y;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    vector<int> ds(n,inf),dt(n,inf);
    queue<int> q;
    ds[s]=0;q.push(s);
    while(q.size()){
        int x=q.front();q.pop();
        for(auto to : v[x]){
            if(ds[to]>ds[x]+1){
                ds[to]=ds[x]+1;
                q.push(to);
            }
        }
    }
    dt[t]=0;q.push(t);
    while(q.size()){
        int x=q.front();q.pop();
        for(auto to : v[x]){
            if(dt[to]>dt[x]+1){
                dt[to]=dt[x]+1;
                q.push(to);
            }
        }
    }
    vector<ll> cnts(n+1),cntt(n+1);
    rep(i,n){
        if(ds[i]<inf)cnts[ds[i]]++;
        if(dt[i]<inf)cntt[dt[i]]++;
    }
    ll ans=0;
    int x = ds[t];
    rep(i,x-1){
        ans+=cnts[i]*cntt[x-2-i];
    }
    cout<<ans<<endl;
    return 0;
}
