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 <vector>
#include <algorithm>
#include <queue>
#include <cstring>


using  namespace std;

#define rep(i,x)for(int i=0;i<(int)(x);++i)

int N, M, s, t;
vector<int> G[100010];
int ds[100010], dt[100010];
int cnts[100010], cntt[100010];

int main()
{
    cin >> N >> M >> s >> t;
    --s, --t;

    rep(i, M) {
        int a, b; cin >> a >> b;
        --a, --b;
        G[a].push_back(b);
        G[b].push_back(a);
    }

    {
        queue<int> q;
        memset(ds, -1, sizeof ds);
        ds[s] = 0;
        q.push(s);
        while (q.size()) {
            int u = q.front(); q.pop();
            for (int v : G[u]) {
                if (ds[v] == -1) {
                    ds[v] = ds[u] + 1;
                    q.push(v);
                }
            }
        }
        rep(i, N) cnts[ds[i]]++;
    }
    {
        queue<int> q;
        memset(dt, -1, sizeof dt);
        dt[t] = 0;
        q.push(t);
        while (q.size()) {
            int u = q.front(); q.pop();
            for (int v : G[u]) {
                if (dt[v] == -1) {
                    dt[v] = dt[u] + 1;
                    q.push(v);
                }
            }
        }
        rep(i, N) cntt[dt[i]]++;
    }

    long long ans = 0;

    int dist = ds[t];

    for (int i = 0; ; ++i) {
        if (i < 0 or dist - i - 2 < 0) break;
        if (1 or i <= dist - i - 2) {
            ans += (long long)cnts[i] * cntt[dist - i - 2];
            //printf("ans += cnts[%d] * cntt[%d]\n", i, dist-i-2);
        }
    }
    
    //rep(i, dist+1) printf("cnts[%d] = %d\n", i, cnts[i]);
    //rep(i, dist+1) printf("cntt[%d] = %d\n", i, cntt[i]);
    

    printf("%lld\n", ans);
}


