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;
unsigned long long positivo = 0, negativo = 0;
void calc(long long a) {
  if (a < 0)
    negativo += (-1 * a);
  else
    positivo += a;
  if (negativo > positivo) {
    negativo -= positivo;
    positivo = 0;
  } else {
    positivo -= negativo;
    negativo = 0;
  }
}
int main() {
  long long a;
  cin >> a;
  long long arr[a];
  for (long long i = 0; i < a; ++i) cin >> arr[i];
  map<long long, long long> n;
  for (long long i = 0; i < a; ++i) {
    calc(-1 * n[arr[i] - 1]);
    calc(n[arr[i] + 1]);
    n[arr[i]]++;
  }
  long long r = 0;
  for (long long i = 1; i < a; i++) {
    r += arr[i] - arr[0];
  }
  calc(r);
  for (long long i = 1; i < a; i++) {
    r += (a - i) * (arr[i - 1] - arr[i]);
    calc(r);
  }
  if (positivo >= negativo)
    cout << positivo << endl;
  else
    cout << '-' << negativo << endl;
  return 0;
}
