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 <queue>
#include <vector>
#include <algorithm>
#include <tuple>
using namespace std;
using P=pair<int,int>;
int main(){
    int N,M,s,t;
    cin>>N>>M>>s>>t;
    s--;t--;
    vector<int> G[N];
    for(int i=0;i<M;i++){
        int x,y;
        cin>>x>>y;
        x--;y--;
        G[x].push_back(y);
        G[y].push_back(x);
    }
    int dist_s[N],dist_t[N];
    fill(dist_s,dist_s+N,1000000000);
    fill(dist_t,dist_t+N,1000000000);
    queue<int> Q;
    dist_s[s]=0;
    Q.push(s);
    while(!Q.empty()){
        int v=Q.front(); Q.pop();
        for(auto to:G[v]){
            if(dist_s[to]==1000000000){
                dist_s[to]=dist_s[v]+1;
                Q.push(to);
            }
        }
    }
    dist_t[t]=0;
    Q.push(t);
    while(!Q.empty()){
        int v=Q.front(); Q.pop();
        for(auto to:G[v]){
            if(dist_t[to]==1000000000){
                dist_t[to]=dist_t[v]+1;
                Q.push(to);
            }
        }
    }
    int64_t cnt_s[N],cnt_t[N];
    fill(cnt_s,cnt_s+N,0);
    fill(cnt_t,cnt_t+N,0);
    for(int i=0;i<N;i++){
        if(dist_s[i]!=1000000000)cnt_s[dist_s[i]]++;
        if(dist_t[i]!=1000000000)cnt_t[dist_t[i]]++;
    }
    int64_t ans=0;
    int dist=dist_s[t];
    for(int i=0;i<=dist-2;i++){
        ans+=cnt_s[i]*cnt_t[dist-2-i];
    }
    cout<<ans<<endl;

    return 0;
}
