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
{
	public static void main(String[] args)
	{
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		long sum=0;
		HashMap<Long,Integer>num=new HashMap<Long,Integer>();
		long[] a=new long[200005];
		for(int i=1;i<=n;i++)
		{
			a[i]=in.nextInt();
			Integer x=num.get(a[i]);
			if(x==null)num.put(a[i],1);
			else num.put(a[i],x+1);
			sum+=a[i];
		}
		BigInteger ans=BigInteger.ZERO;
		int m=n;
		for(int i=1;i<=n;i++)
		{
			Integer xx=num.get(a[i]);
			Integer yy=num.get(a[i]-1);
			Integer zz=num.get(a[i]+1);
			int x,y,z;
			if(xx==null)x=0;else x=xx;
			if(yy==null)y=0;else y=yy;
			if(zz==null)z=0;else z=zz;
			long res=sum-a[i]*x-(a[i]-1)*y-(a[i]+1)*z;
			long cont=m-x-y-z;
			ans=ans.add(BigInteger.valueOf(res-cont*a[i]));
			//System.out.println(sum+" "+res+" "+cont);
			m--;
			sum-=a[i];
			num.put(a[i],x-1);
		}
		System.out.println(ans);
		in.close();
	}
}
