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 D {

	public static void main(String[] args) {
		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 temp = sc.nextInt();
			BigInteger num = BigInteger.valueOf(temp);
			ans = ans.subtract(sum);
			int prev = 0;
			for (int j = -1; j <= 1; j++) {
				if (count.containsKey(temp + j)) {
					prev += count.get(temp + j);
					ans = ans.add(BigInteger.valueOf(count.get(temp + j)).multiply(BigInteger.valueOf(temp + j)));
				}
			}
			ans = ans.add(num.multiply(BigInteger.valueOf(i - prev)));
			if(count.containsKey(temp)){
				count.put(temp, count.get(temp) + 1);
			} else {
				count.put(temp,1);
			}
			
			sum = sum.add(num);
		}
		System.out.println(ans);
	}

}