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. 
import java.math.BigInteger;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
  private static Scanner in = new Scanner(System.in);
  private static final Map<Integer, Integer> mp = new HashMap<>();

  public static void main(String[] args) {
    int n = in.nextInt();
    BigInteger res = BigInteger.ZERO;
    long sum = 0;
    for (int i = 0; i < n; i++) {
      int d = in.nextInt();
      res = res.add(BigInteger.valueOf((long) d * i - sum));
      if (mp.containsKey(d - 1)) {
        int ct = mp.get(d - 1);
        res = res.subtract(BigInteger.valueOf(ct));
      }
      if (mp.containsKey(d + 1)) {
        int ct = mp.get(d + 1);
        res = res.add(BigInteger.valueOf(ct));
      }
      if (mp.containsKey(d)) {
        mp.replace(d, mp.get(d) + 1);
      } else {
        mp.put(d, 1);
      }
      sum += d;
    }
    System.out.println(res);
  }
}