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 pair<int, int> P;
const int M = 1000000007;

int main() {
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    --s;
    --t;
    vector<vector<int>> edge(n);
    for (int i = 0; i < m; ++i) {
        int a, b;
        scanf("%d%d", &a, &b);
        --a;
        --b;
        edge[a].push_back(b);
        edge[b].push_back(a);
    }
    vector<int> ds(n, -1);
    vector<int> cs(n, 0);
    queue<int> q;
    q.push(s);
    ds[s] = 0;
    ++cs[0];
    while (!q.empty()) {
        int p = q.front();
        q.pop();
        for (int i : edge[p]) {
            if (ds[i] == -1) {
                ds[i] = ds[p] + 1;
                ++cs[ds[i]];
                q.push(i);
            }
        }
    }
    vector<int> dt(n, -1);
    vector<int> ct(n, 0);
    q.push(t);
    dt[t] = 0;
    ++ct[0];
    while (!q.empty()) {
        int p = q.front();
        q.pop();
        for (int i : edge[p]) {
            if (dt[i] == -1) {
                dt[i] = dt[p] + 1;
                ++ct[dt[i]];
                q.push(i);
            }
        }
    }
    int l = ds[t] - 1;
    long long ans = 0;
    for (int i = 0; i <= l; ++i) {
        ans += cs[i] * (long long)ct[l - i - 1];
    }
    cout << ans << "\n";
    return 0;
}