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.Scanner;
import java.util.TreeMap;

public class Main {

    TreeMap<Long, Long> cnt = new TreeMap<>();

    public void run() {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();

        BigInteger ans = BigInteger.ZERO;
        BigInteger sum = BigInteger.ZERO;

        for (int i = 0; i < n; i++) {
            long x = in.nextLong();
            long a = cnt.containsKey(x - 1)?cnt.get(x - 1):0;
            long b = cnt.containsKey(x)?cnt.get(x):0;
            long c = cnt.containsKey(x + 1)?cnt.get(x + 1):0;

            ans = ans.add(BigInteger.valueOf(x * (i - a - b - c)));
            ans = ans.add(BigInteger.valueOf(a*(x-1)));
            ans = ans.add(BigInteger.valueOf(b*(x)));
            ans = ans.add(BigInteger.valueOf(c*(x+1)));
            ans = ans.subtract(sum);

            sum = sum.add(BigInteger.valueOf(x));
            if (cnt.containsKey(x)) {
                cnt.put(x, cnt.get(x) + 1);
            } else
                cnt.put(x, 1l);
        }

        System.out.println(ans);
    }

    public static void main(String[] args) {
        new Main().run();
    }
}
