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;

constexpr int INF = 1e9;

vector<int> bfs(vector<vector<int>> const& g, int s) {
    vector<int> d(g.size(), INF);
    d[s] = 0;
    queue<int> que;
    que.push(s);
    while(!que.empty()) {
        int v = que.front();
        que.pop();
        for(auto to : g[v]) {
            if(d[to] == INF) {
                d[to] = d[v] + 1;
                que.push(to);
            }
        }
    }
    return d;
}

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

    auto sd = bfs(g, s), td = bfs(g, t);
    int const min_d = sd[t];
    vector<int> cnt1(N), cnt2(N);
    for(int i = 0; i < N; ++i) {
        if(sd[i] != INF) {
            cnt1[sd[i]]++;
        }
        if(td[i] != INF) {
            cnt2[td[i]]++;
        }
    }
    long long res = 0;
    for(int i = 0; 0 <= min_d - i - 2; ++i) {
        res += 1LL * cnt1[i] * cnt2[min_d - i - 2];
    }
    cout << res << endl;
}
