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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long double n;
  cin >> n;
  vector<long double> v(n);
  for (long double i = 0; i < (long double)n; ++i) {
    cin >> v[i];
  }
  unordered_map<long double, long double> m;
  m[v[0]] = 1;
  long double sum = v[0];
  long double ans = 0;
  for (long double i = 1; i < (long double)n; ++i) {
    long double val = i;
    long double cnt1 = m[v[i]];
    long double val1 = cnt1 * v[i];
    long double cnt2 = m[v[i] - 1];
    long double val2 = cnt2 * (v[i] - 1);
    long double cnt3 = m[v[i] + 1];
    long double val3 = cnt3 * (v[i] + 1);
    long double cnt = cnt1 + cnt2 + cnt3;
    long double ssum = sum - (val1 + val2 + val3);
    ans += (val - cnt) * v[i] - (ssum);
    sum += v[i];
    m[v[i]]++;
  }
  cout << fixed << setprecision(0) << ans << endl;
  return 0;
}
