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. 
//package codeforces.cf903d;

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

public class cf_903_d {
    public static void main(String [] args){
        Scanner cin = new Scanner(System.in);
        int n;
        n = cin.nextInt();
        BigInteger sum = new BigInteger("0");
        sum = BigInteger.valueOf(0L);
        BigInteger ans = new BigInteger("0");
        ans = BigInteger.valueOf(0L);
        HashMap<Long,Long> cnt = new HashMap<Long,Long>();
        for(int i = 1;i <= n;i++){
            long x;
            x = cin.nextLong();
            sum = sum.add(BigInteger.valueOf(x));
            ans = ans.add(BigInteger.valueOf(i * x));
            ans = ans.subtract(sum);
            if(cnt.containsKey(x + 1))
                ans = ans.add(BigInteger.valueOf(cnt.get(x + 1)));
            if(cnt.containsKey(x - 1))
                ans = ans.subtract(BigInteger.valueOf((cnt.get(x-1))));
            if(cnt.containsKey(x)){
                long cur = cnt.get(x);
                cnt.put(x,cur + 1);
            }
            else
                cnt.put(x,1L);
        }
        System.out.println(ans);
    }
}
