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 maxn = 200010;
int n, m, a[maxn], b[maxn], bin[maxn];
long long sum[maxn], ans[maxn];
long double res;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= (int)n; ++i)
    scanf("%d", &a[i]), sum[i] = sum[i - 1] + a[i], b[i] = a[i];
  for (int i = 1; i <= (int)n; ++i) ans[i] = 1ll * a[i] * (i - 1) - sum[i - 1];
  sort(b + 1, b + 1 + n);
  m = unique(b + 1, b + 1 + n) - b - 1;
  for (int i = 1; i <= (int)n; ++i) {
    int tmp = lower_bound(b + 1, b + 1 + m, a[i]) - b;
    if (tmp > 1 && b[tmp - 1] + 1 == b[tmp]) ans[i] -= bin[tmp - 1];
    if (tmp < m && b[tmp + 1] - 1 == b[tmp]) ans[i] += bin[tmp + 1];
    bin[tmp]++;
  }
  for (int i = 1; i <= (int)n; ++i) res = res + ans[i];
  printf("%.0Lf\n", res);
  return 0;
}
