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.io.*;
import java.math.*;
import java.util.*;

public class T5 {

    Random rnd = new Random();
    PrintWriter pw = new PrintWriter(System.out);

    BigInteger gcd(BigInteger x, BigInteger y){
        for(; x.compareTo(BigInteger.ZERO) > 0 && y.compareTo(BigInteger.ZERO)>0;){
            if(x.compareTo(y)>0) x = x.mod(y);
            else y = y.mod(x);
        }
        return x.add(y);
    }

    boolean check_des(BigInteger x){
        for(; x.mod(BigInteger.valueOf(2)).compareTo(BigInteger.ZERO)==0;) x=x.divide(BigInteger.valueOf(2));
        for(; x.mod(BigInteger.valueOf(5)).compareTo(BigInteger.ZERO)==0;) x=x.divide(BigInteger.valueOf(5));
        return x.compareTo(BigInteger.ONE)==0;
    }

    void des(BigInteger x, BigInteger y){
        BigDecimal x1 = new BigDecimal(x.toString());
        BigDecimal y1 = new BigDecimal(y.toString());
        x1 = x1.divide(y1, 50, RoundingMode.CEILING);
        String u = x1.toString();
        for(; u.charAt(u.length()-1)=='0';) u = u.substring(0, u.length()-1);
        pw.print(u);
    }

    void run(){
        Scanner s = new Scanner(System.in);
        int a = s.nextInt();
        long m[] = new long[a];
        for(int q=0; q<a; q++) m[q] = s.nextLong();
        long sum = 0;
        HashMap<Long, Long> hm = new HashMap<>();
        long num[] = new long[a];
        for(int q=a-1; q>=0; q--){
            long c = m[q];

            num[q] = (sum-(hm.getOrDefault(c-1, 0L)*(c-1) + hm.getOrDefault(c, 0L)*c + hm.getOrDefault(c+1, 0L)*(c+1))
                    - c*(a-q-1 - (hm.getOrDefault(c-1, 0L)+hm.getOrDefault(c, 0L)+hm.getOrDefault(c+1, 0L))));

            hm.put(c, hm.getOrDefault(c, 0L)+1);
            sum += c;
        }
        BigInteger o = BigInteger.ZERO;
        for(long u : num) o = o.add(BigInteger.valueOf(u));
        System.out.println(o);
    }

    public static void main(String[] args) {
        new T5().run();
    }
}

