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 long long NN = 111111;
const long long ML = 2147483647;
const long long MLL = 9223372036854775807;
const long double pi = acos(-1);
long long n, m, i, k, j, f, t, l, r, c, e, y, z, w, ii, a[NN], d[NN], p[NN], h1,
    a1, c1, h2, a2;
long double res, s, x;
map<long double, long double> mp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    res += (x * (i - 1 - mp[x - 1] - mp[x + 1]) - s + mp[x - 1] * (x - 1) +
            mp[x + 1] * (x + 1));
    mp[x]++;
    s += x;
  }
  cout << fixed << setprecision(0) << res << "\n";
  return 0;
}
