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>
#pragma comment(linker, "/STACK:167772160000")
using namespace std;
map<long long, long long> totC, c;
long long a[210000];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    totC[a[i]]++;
  }
  long long answer = 0;
  unsigned long long pos, neg;
  pos = neg = 0;
  for (int j = 0; j < (n); j++) {
    long long i = a[j];
    long long lf = c[i - 1] + c[i] + c[i + 1];
    long long rg = totC[i - 1] + totC[i] + totC[i + 1] - 1 - lf;
    lf = j - lf;
    rg = n - j - 1 - rg;
    pos += (unsigned long long)(i * lf);
    neg += (unsigned long long)(i * rg);
    c[i]++;
  }
  if (pos >= neg)
    cout << pos - neg;
  else
    cout << "-" << neg - pos;
  return 0;
}
