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 Main {

    static Map<BigInteger, BigInteger> mp = new HashMap<BigInteger, BigInteger>();

    public static void main(String[] args) {
        mp.clear();
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        BigInteger n = cin.nextBigInteger();
        BigInteger x = cin.nextBigInteger();
        mp.put(x, BigInteger.ONE);
        BigInteger sum = x;
        BigInteger ans = BigInteger.ZERO;
        for (int i = 2;i <= n.intValue(); i++) {
            x=cin.nextBigInteger();
            BigInteger tmp = x.multiply(BigInteger.valueOf(i-1)).subtract(sum);
            if (mp.containsKey(x.subtract(BigInteger.ONE))) tmp = tmp.subtract(mp.get(x.subtract(BigInteger.ONE)));
            if (mp.containsKey(x.add(BigInteger.ONE))) tmp = tmp.add(mp.get(x.add(BigInteger.ONE)));
            ans = ans.add(tmp);
            sum = sum.add(x);
            BigInteger xx;
            if (mp.containsKey(x)) xx = mp.get(x);
            else xx = BigInteger.ZERO;
            mp.put(x, xx.add(BigInteger.ONE));
        }
        System.out.println(ans);
    }
}
