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.*;

public class d {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        Integer[] arr = new Integer[n];
        for (int i = 0; i < n; i++) {
            arr[i] = scanner.nextInt();
        }
        BigInteger minusSum = BigInteger.valueOf(0);
        BigInteger resultSum = BigInteger.valueOf(0);
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            resultSum = resultSum.add(BigInteger.valueOf(arr[i]).multiply(BigInteger.valueOf(i))).subtract(minusSum);
            minusSum = minusSum.add(BigInteger.valueOf(arr[i]));
            map.put(arr[i], map.get(arr[i]) == null ? 1 : map.get(arr[i]) + 1);
            if (map.containsKey(arr[i] - 1)) {
                resultSum = resultSum.subtract(BigInteger.valueOf(map.get(arr[i] - 1)));
            }
            if (map.containsKey(arr[i] + 1)) {
                resultSum = resultSum.add(BigInteger.valueOf(map.get(arr[i] + 1)));
            }
        }

        System.out.println(resultSum);
    }
}