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;
const int N = 200002;
long long n, arr[N], ans, sum[N];
map<long long, long long> mp;
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i)
    cin >> arr[i], sum[i] = sum[i - 1] + arr[i], ++mp[arr[i]];
  long double ans = 0;
  for (int i = 0; i < n; ++i) {
    long long y = sum[n - 1] - sum[i];
    long long a = n - 1 - i;
    ans += (y - a * arr[i]);
    --mp[arr[i]];
    for (int j = -1; j <= 1; ++j) {
      long long x = arr[i] + j;
      a = mp[x];
      ans -= (a * (x - arr[i]));
    }
  }
  cout << fixed << setprecision(0) << ans << endl;
  return 0;
}
