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;

typedef long long ll;

vector<ll>path[101010];
ll depth[2][101010];

ll cnt[2][101010];

int main(){
    for(ll i = 0; i < 101010; i++){
        depth[0][i] = -1;
        depth[1][i] = -1;
    }
    ll N, M, s, t;
    cin >> N >> M >> s >> t;
    for(ll i = 0; i < M; i++){
        ll x, y;
        cin >> x >> y;
        path[x].push_back(y);
        path[y].push_back(x);
    }
    for(ll i = 0; i < 2; i++){
        queue<ll>Q;
        if(i==0){
            Q.push(s);
            depth[i][s] = 0;
        }
        else{
            Q.push(t);
            depth[i][t] = 0;
        }
        while(!Q.empty()){
            ll idx = Q.front();
            Q.pop();
            cnt[i][depth[i][idx]]++;
            for(ll j = 0; j < path[idx].size(); j++){
                ll nxt = path[idx][j];
                if(depth[i][nxt] >= 0)continue;
                depth[i][nxt] = depth[i][idx] + 1;
                Q.push(nxt);
            }
        }
    }
    ll mn = depth[1][s];
    ll ans = 0;
    for(ll a = 0; a <= N; a++){
        ll b = mn - a - 2;
        if(b < 0)continue;
        ans += cnt[0][a] * cnt[1][b];
    }
    cout << ans << endl;
    
    return 0;    
}
