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;
const int N = (int)2e5 + 7;
long long a[N];
map<long long, long long> mp;
int main() {
  long long n;
  scanf("%lld", &n);
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", a + i);
    mp[a[i]]++;
    sum += a[i];
  }
  unsigned long long ans = 0;
  unsigned long long ap = 0, am = 0;
  for (int i = 1; i <= n; i++) {
    long long add =
        (sum - (mp[a[i] - 1] * (a[i] - 1) + mp[a[i]] * a[i] +
                mp[a[i] + 1] * (a[i] + 1))) -
        ((n - i + 1) - mp[a[i] - 1] - mp[a[i]] - mp[a[i] + 1]) * a[i];
    if (add < 0) {
      am = am - add;
    } else {
      ap += add;
    }
    sum -= a[i];
    mp[a[i]]--;
  }
  if (ap >= am) {
    cout << ap - am;
  } else {
    cout << '-' << am - ap;
  }
}
