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;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  long double ans = 0;
  long double a[n + 1];
  map<long double, long double> mp;
  long double sum = 0;
  cout << fixed;
  cout << setprecision(0);
  for (int i = 1; i < n + 1; i++) {
    cin >> a[i];
    mp[a[i]]++;
    long double cp = 0;
    long double cs = 0;
    if (mp.find(a[i] - 1) != mp.end()) cp = mp[a[i] - 1];
    if (mp.find(a[i] + 1) != mp.end()) cs = mp[a[i] + 1];
    long double prev = i - 1;
    prev = prev - cp - cs;
    long double prev_sum = sum - (cp * (a[i] - 1)) - (cs * (a[i] + 1));
    ans = ans + (a[i] * prev) - prev_sum;
    sum = sum + a[i];
  }
  cout << ans << '\n';
  return 0;
}
