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.*;
import java.text.*;

public class Main
{
	static Scanner in = new Scanner(new BufferedInputStream(System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static long a[] = new long[200005];
	static int n;
	
	public static void main(String[]args)
	{
		Map<BigInteger, BigInteger>cnt = new HashMap<BigInteger, BigInteger>();
		int n = in.nextInt();
		for (int i = 1; i <= n; ++i)
		{
			a[i] = in.nextLong();
		}
		BigInteger ans = new BigInteger("0");
		for (int i = 1; i <= n; ++i)
		{
			ans = ans.add(BigInteger.valueOf((i - 1) * a[i]));
			ans = ans.subtract(BigInteger.valueOf((n - i) * a[i]));			
			BigInteger tmp1 = cnt.get(BigInteger.valueOf(a[i] + 1));
			BigInteger tmp2 = cnt.get(BigInteger.valueOf(a[i] - 1));
			if (tmp1 != null)ans = ans.add(tmp1);
			if (tmp2 != null)ans = ans.subtract(tmp2);
			BigInteger tmp3 = cnt.get(BigInteger.valueOf(a[i]));
			if (tmp3 == null)cnt.put(BigInteger.valueOf(a[i]), new BigInteger("1"));
			else cnt.put(BigInteger.valueOf(a[i]), tmp3.add(new BigInteger("1")));

		}
		out.print(ans);
		out.flush();
	}
}