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>
int main() {
  int n;
  std::cin >> n;
  std::map<long long, int> freqs;
  long double common_sum = 0, prev_sum = 0, overall_sum = 0;
  long double curr_number, prev_number;
  std::cin >> curr_number;
  prev_number = curr_number;
  freqs[prev_number] += 1;
  for (int i = 1; i != n; ++i) {
    std::cin >> curr_number;
    common_sum = common_sum + i * (curr_number - prev_number);
    long double a5 = 1 * freqs[curr_number - 1];
    long double a7 = -1 * freqs[curr_number + 1];
    prev_sum = common_sum - a5 - a7;
    freqs[curr_number] += 1;
    prev_number = curr_number;
    overall_sum += prev_sum;
  }
  std::cout << std::fixed << std::setprecision(0) << overall_sum;
}
