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 = 1e6 + 5;
const long long int inf = 1e18 + 5;
long long int a[N], suff[N];
map<long long int, long long int> mp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(0);
  long long int n, t, t1;
  long double sum, ans = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    suff[i] = suff[i + 1] + a[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    t = mp[a[i] + 1] * (a[i] + 1);
    t1 = mp[a[i] - 1] * (a[i] - 1);
    sum = suff[i + 1];
    sum = sum - t - t1 - (mp[a[i]] * a[i]);
    sum = sum - (n - 1 - i - mp[a[i] + 1] - mp[a[i] - 1] - mp[a[i]]) * a[i];
    ans += sum;
    mp[a[i]]++;
  }
  cout << ans;
  return 0;
}
