Let's denote a function 

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You are given an array a consisting of n integers. You have to calculate the sum of d(ai, aj) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.

Input

The first line contains one integer n (1 ≤ n ≤ 200000) — the number of elements in a.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of the array. 

Output

Print one integer — the sum of d(ai, aj) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.

Examples

Input

5
1 2 3 1 3


Output

4


Input

4
6 6 5 5


Output

0


Input

4
6 6 4 4


Output

-8

Note

In the first example:

  1. d(a1, a2) = 0; 
  2. d(a1, a3) = 2; 
  3. d(a1, a4) = 0; 
  4. d(a1, a5) = 2; 
  5. d(a2, a3) = 0; 
  6. d(a2, a4) = 0; 
  7. d(a2, a5) = 0; 
  8. d(a3, a4) = - 2; 
  9. d(a3, a5) = 0; 
  10. d(a4, a5) = 2. 
#include <bits/stdc++.h>
using namespace std;
long long ar[200010], sum[200010];
map<long long, int> mopa;
int main() {
  ios::sync_with_stdio(false);
  int i, j, n, m;
  long long a, b, c = 0, d, x, y, z = 0, sx, sy, sz;
  unsigned long long ans = 0, ans2 = 0;
  cin >> n;
  m = n;
  for (int i = 0; i <= n - 1; i++) {
    cin >> ar[i];
    c += ar[i];
    sum[i] = c;
    mopa[ar[i]]++;
  }
  for (int i = 0; i <= n - 1; i++) {
    x = mopa[ar[i] - 1];
    y = mopa[ar[i]] - 1;
    z = mopa[ar[i] + 1];
    sx = x * (ar[i] - 1);
    sy = y * (ar[i]);
    sz = z * (ar[i] + 1);
    a = m - (x + y + z);
    d = c - (sx + sy + sz);
    if ((d - a * ar[i]) > 0)
      ans += (d - a * ar[i]);
    else
      ans2 += (-1 * (d - a * ar[i]));
    m--;
    c -= ar[i];
    mopa[ar[i]]--;
  }
  if (ans >= ans2)
    cout << ans - ans2 << '\n';
  else
    cout << '-' << ans2 - ans << '\n';
  return 0;
}
