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.util.*;
import java.math.*;
public class Main
{
	public static void main(String[] args)
	{
		Scanner cin=new Scanner(System.in);
		int n=cin.nextInt();
		HashMap<Integer,Integer>cnt=new HashMap<Integer,Integer>();
		BigInteger ans=BigInteger.ZERO;
		BigInteger sum=BigInteger.ZERO;
		for(int i=0;i<n;++i)
		{
			int x=cin.nextInt();
			BigInteger num=BigInteger.valueOf(i);
			BigInteger al=sum;
			for(int j=-1;j<=1;++j)
			{
				int v=x+j;
				if(cnt.containsKey(v)) 
				{	
					int tim=cnt.get(v);
					num=num.subtract(BigInteger.valueOf(tim));
					al=al.subtract(BigInteger.valueOf(tim).multiply(BigInteger.valueOf(v)));
				}
			}
			ans=ans.add(num.multiply(BigInteger.valueOf(x)));
			ans=ans.subtract(al);
			if(cnt.containsKey(x)) cnt.put(x,cnt.get(x)+1);
			else cnt.put(x, 1);
			sum=sum.add(BigInteger.valueOf(x));
		}
		
		System.out.println(ans);
	}
}