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 = 2e5 + 10, MOD = 1e9 + 7, MAXN = 1e5 + 10;
const long double inf = 1e10 + 10;
long double a[N];
map<long double, long double> m;
int main() {
  int n;
  cin >> n;
  long double sum = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long double ans = 0;
  sum += a[n];
  m[a[n]]++;
  int j = 1;
  for (int i = n - 1; i > 0; i--) {
    long double cnt = sum;
    cnt -= m[a[i]] * a[i];
    cnt -= m[a[i] + 1] * (a[i] + 1);
    cnt -= m[a[i] - 1] * (a[i] - 1);
    long double tmp = m[a[i]] + m[a[i] + 1] + m[a[i] - 1];
    ans += cnt - (j - tmp) * a[i];
    sum += a[i];
    m[a[i]]++;
    j++;
  }
  cout << fixed << setprecision(0) << ans << endl;
  return 0;
}
