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;
map<long double, int> cant;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<long double> A(n + 1);
  for (int i = (1); i < (n + 1); i++) cin >> A[i];
  vector<long double> prefixSum(n + 2);
  prefixSum[0] = 0;
  for (int i = (1); i < (n + 2); i++) {
    prefixSum[i] = prefixSum[i - 1] + A[i];
  }
  long double total = 0;
  for (int i = (1); i < (n + 1); i++) {
    long double prim = prefixSum[i - 1] - A[i] * (i - 1);
    long double sec = A[i] * cant[A[i] - 1.] - (A[i] - 1.) * cant[A[i] - 1.];
    long double tri = A[i] * cant[A[i] + 1.] - (A[i] + 1.) * cant[A[i] + 1.];
    total -= (prim + sec + tri);
    cant[A[i]]++;
  }
  cout << fixed << setprecision(0) << total << "\n";
}
