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 Main {
    public static void main(String[] args){
        int a[] = new int [202000];
        HashMap<Integer, Integer> mp = new HashMap<>();
        int n;
        Scanner in;
        in = new Scanner(System.in);
        n = in.nextInt();
        BigInteger ans = BigInteger.ZERO;
        long sum = 0;
        for(int i=1; i<=n; i++) {
            int u = in.nextInt();
            long s = sum;
            int cnt = i - 1;
            for(int j=-1; j<=1; j++) {
                if(mp.containsKey(u+j)) {
                    s -= (long) mp.get(u+j) * (u + j);
                    cnt -= mp.get(u+j);
                }
            }
            if(mp.containsKey(u)) mp.replace(u, mp.get(u)+1);
            else mp.put(u, 1);
            sum += u;
            ans = ans.add(BigInteger.valueOf((long) u * cnt - s));
        }
        System.out.print(ans);
    }
}

                             