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 javax.swing.*;
import java.math.BigInteger;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class AlmostDifference {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int ar[] = new int[n];
        Map<Integer, Integer> mp = new TreeMap<>();
        BigInteger sum = BigInteger.valueOf(0);
        for (int i = 0; i < n; i++){
            ar[i] = sc.nextInt();
            sum = sum.add(BigInteger.valueOf(ar[i]));
            mp.put(ar[i], mp.get(ar[i]) == null ? 1 : mp.get(ar[i]) + 1);
        }
        BigInteger ans = BigInteger.ZERO;
        BigInteger el = BigInteger.valueOf(n);
        for (int i = 0; i < n; i++){
            BigInteger bad = BigInteger.ZERO;
            BigInteger bad_sum = BigInteger.ZERO;
            for (int j = -1; j <= 1; j++){
                int val = (ar[i] + j);
                long bb = mp.get(val) == null ? 0 : mp.get(val);
                bad = bad.add(BigInteger.valueOf(bb));
                long bs = bb * (long) val;
                bad_sum = bad_sum.add(BigInteger.valueOf(bs));
            }
            BigInteger good = (el.subtract(bad));
            BigInteger curr = (sum.subtract(bad_sum));
            BigInteger tmp = curr.subtract(good.multiply(BigInteger.valueOf(ar[i])));
            ans = ans.add(tmp);
            sum = sum.subtract(BigInteger.valueOf(ar[i]));
            el = el.subtract(BigInteger.ONE);
            mp.put(ar[i], mp.get(ar[i]) - 1);
        }
        System.out.println(ans.toString());
    }
}
