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.Scanner;
import java.util.Map;

public final class Main {
    public static void main(String args[]) {
        Scanner scanner = new Scanner(System.in);

        long n;
        n = scanner.nextLong();

        Map< Long , Long >mp = new HashMap<>();
        BigInteger ans = new BigInteger("0");
        long cu = 0;

        for (int i = 1; i <= n; i++) {
            long a = scanner.nextLong();

            if (mp.containsKey(a)) mp.put(a, mp.get(a)+1);
            else mp.put(a, 1L);

            cu += a;

            long x = mp.get(a), y = 0, z = 0;
            if (mp.containsKey(a-1)) y = mp.get(a-1);
            if (mp.containsKey(a+1)) z = mp.get(a+1);

            long tmp = ((i-x-y-z)*a-(cu-a*x-(a-1)*y-(a+1)*z));

            //System.out.println(tmp);

            ans = ans.add(new BigInteger(Long.toString(tmp)));
        }

        System.out.println(ans);
    }
}

