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>
#define REP(i, a, n) for(int i = ((int) a); i < ((int) n); i++)
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;

ll N, M, S, T, X, Y;
vector<ll> E[100010];

ll wfs(ll *count, ll s, ll t) {
  bool visited[100010];
  REP(i, 0, N) visited[i] = false;
  REP(i, 0, N) count[i] = 0;

  ll dist;
  queue<pii> q;
  q.push(pii(0, s));

  while(q.size()) {
    ll d = q.front().first, v = q.front().second;
    q.pop();

    if(visited[v]) continue;
    visited[v] = true;

    count[d]++;

    if(v == t) {
      dist = d;
      break;
    }

    REP(i, 0, E[v].size()) if(!visited[E[v][i]]) q.push(pii(d + 1, E[v][i]));
  }

  return dist;
}

int main(void) {
  cin >> N >> M >> S >> T; S--; T--;
  REP(i, 0, M) {
    cin >> X >> Y; X--; Y--;
    E[X].push_back(Y);
    E[Y].push_back(X);
  }

  ll scount[100010], tcount[100010];
  ll sdist = wfs(scount, S, T);
  ll tdist = wfs(tcount, T, S);

  ll ans = 0;
  REP(i, 0, sdist - 1) ans += scount[i] * tcount[tdist - i - 2];
  cout << ans << endl;

  return 0;
}