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;
unordered_map<long long, long long> mymap;
long long sum = 0;
long double ans;
long long arr[1000005];
int main() {
  long long n;
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", arr + i);
    mymap[arr[i]]++;
    sum = sum + arr[i];
  }
  long long tmpn = n;
  for (int i = 0; i < n; i++) {
    long long pv = arr[i] - 1;
    long long cv = arr[i];
    long long nv = arr[i] + 1;
    long long cpv = mymap[pv];
    long long ccv = mymap[cv];
    long long cnv = mymap[nv];
    long long value = sum - ((cpv * pv) + (cnv * nv) + (cv * ccv));
    long long tot = tmpn - (cpv + ccv + cnv);
    value = value - (tot * arr[i]);
    ans = ans + (long double)value;
    mymap[arr[i]]--;
    sum = sum - arr[i];
    tmpn--;
  }
  cout << fixed << setprecision(0) << ans;
}
