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 N = 2e5 + 5;
long long n, a[N];
map<long long, long long> cnt, cnt1;
long long get(long long x) {
  if (cnt.find(x) == end(cnt))
    return 0;
  else
    return cnt[x];
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 1, _ = n; i <= _; ++i) cin >> a[i];
  long double tot = 0;
  long long s = 0;
  for (long long i = 1, _ = n; i <= _; ++i) cnt[a[i]]++;
  for (long long i = 1, _ = n; i <= _; ++i) {
    tot += (i - 1) * a[i] - (n - i) * a[i];
    tot += cnt[a[i] - 1];
    tot -= cnt1[a[i] - 1];
    cnt[a[i]]--;
    cnt1[a[i]]++;
  }
  cout << setprecision(0) << fixed;
  cout << tot << '\n';
}
