Let's denote a function 

<image>

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 long int, long long int> present;
map<long long int, long long int> running;
int main() {
  long long int a;
  cin >> a;
  long long int num[a], sum[a], rfreq, minus, ans, sign = 1;
  unsigned long long int ptotal = 0, ntotal = 0;
  for (long long int d = 0; d < a; d++) {
    cin >> num[d];
    if (d == 0) {
      sum[d] = num[d];
    } else {
      sum[d] = sum[d - 1] + num[d];
    }
    present[num[d]]++;
  }
  for (long long int d = 0; d < a; d++) {
    running[num[d]]++;
    minus = 0, rfreq = 0, ans = 0;
    if (present[num[d] + 1] > 0) {
      minus += ((num[d] + 1) * (present[num[d] + 1] - running[num[d] + 1]));
      rfreq += (present[num[d] + 1] - running[num[d] + 1]);
    }
    if (present[num[d]] > 1) {
      minus += ((num[d]) * (present[num[d]] - running[num[d]]));
      rfreq += (present[num[d]] - running[num[d]]);
    }
    if (present[num[d] - 1] > 0) {
      minus += ((num[d] - 1) * (present[num[d] - 1] - running[num[d] - 1]));
      rfreq += (present[num[d] - 1] - running[num[d] - 1]);
    }
    ans = ((sum[a - 1] - sum[d]) - minus - num[d] * (a - d - 1 - rfreq));
    if (ans < 0) {
      ans *= -1;
      ntotal = ntotal + ans;
    } else {
      ptotal = ptotal + ans;
    }
  }
  if (ptotal >= ntotal) {
    cout << ptotal - ntotal;
  } else {
    cout << "-" << ntotal - ptotal;
  }
}
