Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.

In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease. 

You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won't decrease.

Input

The firt line of the input contains integers n, m, s and t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.

Output

Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won't decrease the distance between junctions s and t.

Examples

Input

5 4 1 5
1 2
2 3
3 4
4 5


Output

0


Input

5 4 3 5
1 2
2 3
3 4
4 5


Output

5


Input

5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5


Output

3
#include <bits/stdc++.h>
using namespace std;
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<long long int> prime;
void sieve(long long int n) {
  bool bakh[n + 1];
  memset(bakh, true, sizeof(bakh));
  for (long long int p = 2; p * p <= n; p++) {
    if (bakh[p] == true) {
      for (long long int i = p * p; i <= n; i += p) bakh[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (bakh[p]) prime.push_back(p);
}
long long int eulertotient(long long int z) {
  long long int fac = z;
  for (long long int i = 0; prime[i] * prime[i] <= z; i++) {
    if (z % prime[i] == 0) {
      fac -= (fac / prime[i]);
      while (z % prime[i] == 0) z /= prime[i];
    }
  }
  if (z > 1) fac -= (fac / z);
  return fac;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  long long int g = gcd(a, b);
  long long int ans = (a * b) / g;
  return ans;
}
long long int modInverse(long long int a, long long int m) {
  long long int hvf = gcd(a, m);
  if (hvf == 1) return power(a, m - 2, m);
  return -1;
}
void multiply(long long int F[2][2], long long int M[2][2]) {
  long long int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
  long long int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
  long long int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
  long long int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
void powermat(long long int F[2][2], long long int n) {
  if (n == 0 || n == 1) return;
  long long int M[2][2] = {{1, 1}, {1, 0}};
  powermat(F, n / 2);
  multiply(F, F);
  if (n % 2 != 0) multiply(F, M);
}
long long int fib(long long int n) {
  long long int F[2][2] = {{1, 1}, {1, 0}};
  if (n == 0) return 0;
  powermat(F, n - 1);
  return F[0][0];
}
long long int n, m, s, t;
long long int ds[1005], dt[1005];
long long int mat[1005][1005];
bool mark[1005];
vector<long long int> adj[1005];
void bfs1(long long int v) {
  queue<pair<long long int, long long int> > q;
  q.push(make_pair(v, 0));
  ds[v] = 0;
  mark[v] = true;
  while (!q.empty()) {
    pair<long long int, long long int> a = q.front();
    q.pop();
    long long int p = a.first;
    long long int dis = a.second;
    ds[p] = dis;
    for (long long int j = 0; j < adj[p].size(); j++) {
      if (!mark[adj[p][j]])
        q.push(make_pair(adj[p][j], dis + 1)), mark[adj[p][j]] = true;
    }
  }
}
void bfs2(long long int v) {
  queue<pair<long long int, long long int> > q;
  q.push(make_pair(v, 0));
  dt[v] = 0;
  mark[v] = true;
  while (!q.empty()) {
    pair<long long int, long long int> a = q.front();
    q.pop();
    long long int p = a.first;
    long long int dis = a.second;
    dt[p] = dis;
    for (long long int j = 0; j < adj[p].size(); j++) {
      if (!mark[adj[p][j]])
        q.push(make_pair(adj[p][j], dis + 1)), mark[adj[p][j]] = true;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m >> s >> t;
  long long int a, b;
  for (long long int i = 0; i < m; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    mat[a][b] = 1;
    mat[b][a] = 1;
  }
  bfs1(s);
  memset(mark, false, sizeof(mark));
  bfs2(t);
  long long int ans = 0, dis;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= n; j++) {
      if (i == j || mat[i][j] == 1) continue;
      dis = min(ds[i] + dt[j] + 1, ds[j] + dt[i] + 1);
      if (dis >= ds[t]) ans++;
    }
  }
  cout << ans / 2 << endl;
}
