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 Main {
	static PrintStream out = new PrintStream(System.out);
	static Scanner scan = new Scanner(System.in);
	static int n, a;
	static int[] arr;
	static HashMap<Integer, Integer> map;
	static BigInteger res, sum;
	
	
	public static void main (String[] args) {
		res = BigInteger.ZERO;
		sum = BigInteger.ZERO;
		
		n = scan.nextInt();
		map = new HashMap<Integer, Integer>();
		
		for (int i = 1; i <= n; i++) {
			a = scan.nextInt();
			BigInteger bigA = BigInteger.valueOf(a);
			
			sum = sum.add(BigInteger.valueOf(a));
			res = res.subtract(sum);
			res = res.add(bigA.multiply(BigInteger.valueOf(i)));
			res = res.subtract(BigInteger.valueOf(get(a - 1)));
			res = res.add(BigInteger.valueOf(get(a + 1)));
			put(a);
		}
		out.println(res.toString());
	}
	
	private static int get (int a) {
		if (map.containsKey(a)) return map.get(a);
		return 0;
	}
	
	private static void put (int a) {
		if (map.containsKey(a)) map.put(a, map.get(a).intValue() + 1);
		else map.put(a, 1);
	}
}