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 sc = new Scanner(System.in);
		int n;
		n = sc.nextInt();
		int a[] = new int[n + 1];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		BigInteger sum = BigInteger.valueOf(0);
		HashMap<Integer, Integer> visit = new HashMap<>();
		for (int i = 0; i < n; i++) {
			long j = 1;
			sum = sum.add(BigInteger.valueOf(-j * (n - i - 1) * a[i]));
			sum = sum.add(BigInteger.valueOf(j * i * a[i]));
			if (visit.containsKey(a[i] - 1)) {
				sum = sum.add(BigInteger.valueOf(-visit.get(a[i] - 1)));
			}
			if (visit.containsKey(a[i] + 1)) {
				sum = sum.add(BigInteger.valueOf(visit.get(a[i] + 1)));
			}
			if (visit.containsKey(a[i]) == false) {
				visit.put(a[i], 1);
			} else {
				visit.put(a[i], visit.get(a[i]) + 1);
			}

		}
		System.out.println(sum);
		sc.close();
	}

}
