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;

typedef long long ll;

vector< vector<int> > G;
int dist[2][100100];
ll cnt[2][100100];
int N, M, s, t;

void bfs(int v)
{
  int c = v == s ? 0 : 1;
  queue<int> que;
  dist[c][v] = 0;
  que.push(v);
  while(!que.empty()) {
    int now = que.front(); que.pop();
    for(int i = 0; i < G[now].size(); i++) {
      int to = G[now][i];
      if(!~dist[c][to]) {
	dist[c][to] = dist[c][now] + 1;
	que.push(to);
      }
    }
  }
}

int main()
{
  cin >> N >> M >> s >> t;
  --s, --t;
  G.resize(N);
  for(int i = 0; i < M; i++) {
    int x, y; cin >> x >> y;
    --x, --y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  memset(dist, -1, sizeof(dist));
  memset(cnt, 0, sizeof(cnt));  
  bfs(s); bfs(t);  
  for(int i = 0; i < N; i++) {
    cnt[0][dist[0][i]]++;
    cnt[1][dist[1][i]]++;
  }
  ll ans = 0LL;
  for(int i = 0; i < dist[0][t]-1; i++) {
    ans += cnt[0][i]*cnt[1][dist[0][t]-i-2];
  }
  cout << ans << endl;
  return 0;
}