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.HashMap;
import java.util.Scanner;

public class Main {

	static Long correct(Long g){
		if(g == null) return 0L;
		else return g;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		HashMap<Long, Long> mp = new HashMap<>();
		int n = sc.nextInt();
		BigInteger ans = BigInteger.ZERO, s = BigInteger.ZERO;
		for(int i = 0; i < n; i++){
			Long x = sc.nextLong();
			Long f = 1L * i * 1L * x;
			f += 1L * correct(mp.get(x + 1));
			f -= 1L * correct(mp.get(x - 1));
			ans = ans.add(new BigInteger(f.toString()).subtract(s));
			s = s.add(new BigInteger(x.toString()));
			Long y = correct(mp.get(x));
			mp.put(x, y + 1);
		}
		
		System.out.println(ans);
	}
}