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;

using Edges=vector<int>;
using Graph=vector<Edges>;
using ll=long long;
const int INF=1e9;

void make_dist(int v,vector<int> &d,const Graph &g){
    d[v]=0;
    queue<int> que;
    que.push(v);
    while(!que.empty()){
        int ver=que.front(); que.pop();
        for(int i=0;i<g[ver].size();i++){
            int to=g[ver][i];
            if(d[ver]+1<d[to]){
                d[to]=d[ver]+1;
                que.push(to);
            }
        }
    }
}
int main(){
    int n,m,s,t;
    cin>>n>>m>>s>>t;
    s--,t--;
    Graph 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);
    }
    vector<int> ds(n,INF);
    make_dist(s,ds,g);
    vector<int> dt(n,INF);
    make_dist(t,dt,g);

    map<int,ll> cnts;
    map<int,ll> cntt;
    for(int i=0;i<n;i++){
        cnts[ds[i]]++;
        cntt[dt[i]]++;
    }

    int dist=ds[t];
    ll res=0;
    for(auto &e:cnts){
        res+=e.second*cntt[dist-e.first-2];
    }
    cout<<res<<endl;
    return 0;
}
