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 REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)

using namespace std;

typedef long long ll;

const int MAX = 100010;
const int IINF = INT_MAX;
int V,E,s,t;
vector<int> G[MAX];
int mindist[2][MAX]; // 0 -> start, 1 -> goal
ll counter[2][MAX];

void BFS(int sp){
  int index = (sp==t);
  mindist[index][sp] = 0;
  deque<int> deq;
  deq.push_back(sp);
  while( !deq.empty() ){
    int cur = deq.front(); deq.pop_front();
    rep(i,G[cur].size()){
      int next = G[cur][i];
      if( mindist[index][next] == IINF ) {
        mindist[index][next] = mindist[index][cur] + 1;
        deq.push_back(next);
      }
    }
  }
}

void compute(){
  rep(i,2) rep(j,V) mindist[i][j] = IINF, counter[i][j] = 0;
  BFS(s);
  BFS(t);
  int mini = mindist[0][t];
  ll answer = 0;
  rep(i,V) {
    if( mindist[0][i] != IINF ) ++counter[0][mindist[0][i]];
    if( mindist[1][i] != IINF ) ++counter[1][mindist[1][i]];
  }
  rep(i,mini-1) answer += ( counter[0][i] * counter[1][mini-2-i] );
  cout << answer << endl;
}

int main(){
  scanf("%d %d %d %d",&V,&E,&s,&t);
  --s,--t;
  rep(i,E) {
    int a,b;
    scanf("%d %d",&a,&b);
    --a, --b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  compute();
  return 0;
}