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 = 2e5 + 5;
int n;
long long sum, a[N];
unsigned long long pos, neg;
map<long long, int> c;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = n; i; i--) {
    long long _sum = sum;
    int con = n - i;
    for (long long j = a[i] - 1; j <= a[i] + 1; j++) {
      _sum -= 1ll * c[j] * j;
      con -= c[j];
    }
    long long now = _sum - 1ll * con * a[i];
    if (now >= 0) {
      pos += now;
    } else {
      neg += -now;
    }
    c[a[i]]++;
    sum += a[i];
  }
  if (pos >= neg) {
    cout << pos - neg << "\n";
  } else {
    cout << "-" << neg - pos << "\n";
  }
}
