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;

#define lint long long
#define P pair<int, int>
#define LLP pair<long long, long long>
#define REP(i, x, n) for(int i = (x), i##_len = (int)(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = (int)(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = (int)(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = (int)(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = (int)(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())

const int IINF = (1 << 30) - 1;
const long long LLINF = 1LL << 61;
const long long MOD = 1000000007LL;
const int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};
const double EPS = 1e-8;

template<typename T>
bool chmin(T &_a, T _b){
    if(_a > _b){
        _a = _b;
        return true;
    }else{
        return false;
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, m;
    vector<int> st(2);
    cin >> n >> m >> st[0] >> st[1];
    rep(i, 2){
        --st[i];
    }

    vector< vector<int> > g(n);
    rep(i, m){
        int u, v;
        cin >> u >> v;
        --u;
        --v;
        g[u].emplace_back(v);
        g[v].emplace_back(u);
    }

    vector< vector<int> > cost(2, vector<int>(n, IINF));
    rep(i, 2){
        queue< P > que;
        que.push({st[i], 0});
        while(!que.empty()){
            P now = que.front();
            que.pop();
            if(!chmin(cost[i][now.first], now.second)){
                continue;
            }
            for(auto e : g[now.first]){
                if(cost[i][e] > now.second + 1){
                    que.push({e, now.second + 1});
                }
            }
        }
    }

    int d = cost[0][st[1]];

    vector< vector<lint> > cnt(2, vector<lint>(d + 1, 0LL));
    rep(i, 2){
        rep(j, n){
            if(cost[i][j] <= d){
                ++cnt[i][cost[i][j]];
            }
        }
    }

    lint ans = 0;
    rep(i, d - 1){
        ans += cnt[0][i] * cnt[1][d - i - 2];
    }

    cout << ans << endl;

    return 0;
}
