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

public class D {
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		
		BigInteger ans = BigInteger.ZERO;
		BigInteger sum = BigInteger.ZERO;
		
		HashMap<Integer,Integer> count = new HashMap<Integer,Integer>();
		
		for(int i = 0; i < N; i++) {
			int in = sc.nextInt();
			BigInteger a = BigInteger.valueOf(in);
			
			ans = ans.subtract(sum);
		
			int prev = 0;
			for(int j = -1; j <= 1; j++) {
				if(count.containsKey(in+j)) {
					prev += count.get(in+j);
					ans = ans.add(BigInteger.valueOf(count.get(in+j)).multiply(BigInteger.valueOf(in+j)));
				}
			}
			
			ans = ans.add(a.multiply(BigInteger.valueOf(i-prev)));
			
			if(count.containsKey(in)) count.put(in,count.get(in)+1);
			else count.put(in,1);
		
			sum = sum.add(a);
		}		
		
		System.out.println(ans);
	}
	
}
