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>
#define For(i, a, b) for(int (i)=(int)(a); (i)<(int)(b); ++(i))
#define rFor(i, a, b) for(int (i)=(int)(a)-1; (i)>=(int)(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef unsigned long long ulint;
typedef pair<int, int> pii;
typedef pair<lint, lint> pll;
template<class T> bool chmax(T &a, const T &b){if(a<b){a=b; return true;} return false;}
template<class T> bool chmin(T &a, const T &b){if(a>b){a=b; return true;} return false;}
template<class T> T div_floor(T a, T b){
    if(b < 0) a *= -1, b *= -1;
    return a>=0 ? a/b : (a+1)/b-1;
}
template<class T> T div_ceil(T a, T b){
    if(b < 0) a *= -1, b *= -1;
    return a>0 ? (a-1)/b+1 : a/b;
}

constexpr lint mod = 1000000007;
constexpr lint INF = mod * mod;
constexpr int MAX = 100010;

int n, m;
vector<int> G[MAX];

vector<int> bfs(int s){
    vector<int> d(n, mod);
    d[s] = 0;
    queue<int> que;
    que.push(s);
    while(!que.empty()){
        int v = que.front();
        que.pop();
        for(auto nv: G[v])if(chmin(d[nv], d[v] + 1)){
            que.push(nv);
        }
    }
    return d;
}

int main(){
    int s, t;
    scanf("%d%d%d%d", &n, &m, &s, &t);
    --s; --t;
    rep(i, m){
        int a, b;
        scanf("%d%d", &a, &b);
        --a; --b;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    auto ds = bfs(s), dt = bfs(t);
    int dst = ds[t];
    lint ans = 0, cnt[n+10];
    memset(cnt, 0LL, sizeof(cnt));
    rep(i, n)if(dt[i] < n+10) ++cnt[dt[i]];
    rep(i, n)if(dst-2 - ds[i] >= 0) ans += cnt[dst-2 - ds[i]];
    printf("%lld\n", ans);
}
