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 a[N];
map<int, int> mp;
long long suf[N];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = n; i >= 1; --i) {
    suf[i] = suf[i + 1] + a[i];
  }
  long double sum = 0;
  for (int i = n; i >= 1; --i) {
    int x = mp[a[i]];
    int y = mp[a[i] - 1];
    int z = mp[a[i] + 1];
    sum += suf[i + 1] - 1ll * x * a[i] - 1ll * y * (a[i] - 1) -
           1ll * z * (a[i] + 1) - 1ll * (n - i - x - y - z) * a[i];
    ++mp[a[i]];
  }
  cout << fixed << setprecision(0) << sum;
}
