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;

int main() {
    int N, M;
    cin >> N >> M;
    vector<bool> isReached(N, false), is(N,false);
    vector<ll> count(N,0),c(N,0);
    vector< vector<int> > graph(N);
    queue<int> que;
    vector<int> dist(N),d(N);
    int s, t;
    cin >> s >> t;
    s--; t--;
    int x, y;
    for(int i=0 ; i<M ; i++){
        cin >> x >> y;
        x--; y--;
        graph[x].push_back(y);
        graph[y].push_back(x);
    }
    int now;
    que.push(s);
    count[0] = 1;
    isReached[s] = true;
    dist[s] = 0;
    while(!que.empty()){
        now = que.front();
        que.pop();
        for(int i=0; i<graph[now].size() ; i++){
            if(!isReached[graph[now][i]]){
                que.push(graph[now][i]);
                isReached[graph[now][i]] = true;
                dist[graph[now][i]] = dist[now]+1;
                count[dist[graph[now][i]]]++;
            }
        }
    }
    que.push(t);
    c[0] = 1;
    is[t] = true;
    d[t]=0;
    while(!que.empty()){
        now = que.front();
        que.pop();
        for(int i=0; i<graph[now].size() ; i++){
            if(!is[graph[now][i]]){
                que.push(graph[now][i]);
                is[graph[now][i]] = true;
                d[graph[now][i]] = d[now]+1;
                c[d[graph[now][i]]]++;
            }
        }
    }
    ll ans=0;
    for(int i=0; i<= dist[t]-2 ; i++){
        ans+= count[i]*c[dist[t]-2-i];
    }
    cout << ans << endl;
    return 0;
}

