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>
#pragma GCC target("sse4,avx")
void run(std::istream& in, std::ostream& out) {
  int n;
  in >> n;
  std::vector<int> as(n);
  for (int i = 0; i < n; i++) {
    in >> as[i];
  }
  std::map<int, int> counts;
  int64_t sum = 0;
  uint64_t pos = 0;
  uint64_t neg = 0;
  for (int i = n - 1; i >= 0; i--) {
    pos += sum;
    neg += int64_t(as[i]) * (n - i - 1);
    neg += counts[as[i] + 1];
    pos += counts[as[i] - 1];
    uint64_t d = std::min(pos, neg);
    pos -= d;
    neg -= d;
    counts[as[i]]++;
    sum += as[i];
  }
  if (pos >= neg) {
    out << pos - neg << "\n";
  } else {
    out << "-" << neg - pos << "\n";
  }
}
int main() {
  std::cin.sync_with_stdio(false);
  std::cin.tie(nullptr);
  run(std::cin, std::cout);
  return 0;
}
