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 <queue>

using namespace std;
using ll = long long;

struct Graph {
    Graph(const int n)
    {
        edge.resize(n);
    }
    void addEdge(const int from, const int to)
    {
        edge[from].push_back(to);
    }
    vector<vector<int>> edge;
};


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.addEdge(x, y);
        g.addEdge(y, x);
    }
    vector<bool> used(N, false);
    using P = pair<int, int>;
    queue<P> q;
    used[s] = true;
    q.push(make_pair(s, 0));
    constexpr int INF = 1 << 30;
    vector<int> dist(N, INF);
    while (not q.empty()) {
        const P p = q.front();
        dist[p.first] = p.second;
        q.pop();
        for (const int to : g.edge[p.first]) {
            if (not used[to]) {
                used[to] = true;
                q.push(make_pair(to, p.second + 1));
            }
        }
    }
    while (not q.empty()) {
        q.pop();
    }

    fill(used.begin(), used.end(), false);
    used[t] = true;
    q.push(make_pair(t, 0));
    vector<int> back_dist(N, INF);
    while (not q.empty()) {
        const P p = q.front();
        back_dist[p.first] = p.second;
        q.pop();
        for (const int to : g.edge[p.first]) {
            if (not used[to]) {
                used[to] = true;
                q.push(make_pair(to, p.second + 1));
            }
        }
    }
    const int D = dist[t] - 2;

    vector<ll> num(D + 1, 0);
    vector<ll> back_num(D + 1, 0);
    for (int i = 0; i < N; i++) {
        if (dist[i] <= D) {
            num[dist[i]]++;
        }
        if (back_dist[i] <= D) {
            back_num[back_dist[i]]++;
        }
    }

    ll sum = 0;
    for (int i = 0; i <= D; i++) {
        sum += num[i] * back_num[D - i];
    }
    cout << sum << endl;

    return 0;
}