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;
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  map<int, int> prev;
  map<int, int> aft;
  int vals[n];
  unsigned long long ansp = 0LL;
  unsigned long long ansn = 0LL;
  for (int i = 0; i < n; i++) {
    cin >> vals[i];
    prev[vals[i]] = 0;
    aft[vals[i]] = 0;
  }
  for (int i = 0; i < n; i++) {
    aft[vals[i]]++;
  }
  for (int i = 0; i < n; i++) {
    int spot = i;
    aft[vals[i]]--;
    spot -= prev[vals[i]];
    if (prev.count(vals[i] - 1) > 0) {
      spot -= prev[vals[i] - 1];
    }
    if (prev.count(vals[i] + 1) > 0) {
      spot -= prev[vals[i] + 1];
    }
    ansp += (spot + 0LL) * (vals[i] + 0LL);
    spot = n - i - 1;
    spot -= aft[vals[i]];
    if (aft.count(vals[i] - 1) > 0) {
      spot -= aft[vals[i] - 1];
    }
    if (aft.count(vals[i] + 1) > 0) {
      spot -= aft[vals[i] + 1];
    }
    ansn += (spot + 0LL) * (vals[i] + 0LL);
    prev[vals[i]]++;
  }
  if (ansp >= ansn) {
    ansp -= ansn;
    cout << ansp << endl;
  } else {
    ansn -= ansp;
    cout << "-" << ansn << endl;
  }
  cin >> ansp;
}
