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.Scanner;
import java.util.TreeMap;

public class D_Almost_Difference {

	@SuppressWarnings("resource")
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		TreeMap<Long, Long> record = new TreeMap<Long, Long>();
		long prevSum = 0;
		BigInteger ans = BigInteger.ZERO;
		for (int i = 0; i < n; i++){
			long a = input.nextInt();
			long bc = i;
			long bt = prevSum;
			if (record.containsKey(a - 1)){
				bc -= record.get(a - 1);
				bt -= record.get(a - 1) * (a - 1);
			}
			if (record.containsKey(a + 1)){
				bc -= record.get(a + 1);
				bt -= record.get(a + 1) * (a + 1);
			}
			if (record.containsKey(a)){
				bc -= record.get(a);
				bt -= record.get(a) * (a);
				record.put(a, record.get(a) + 1);
			}else{
				record.put(a, 1L);
			}
			//System.out.println(bc + " " + bt);
			ans = ans.add(new BigInteger(""+(bc * a - bt)));
			prevSum += a;
		}
		System.out.println(ans);
	}

}
