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.io.PrintWriter;
import java.math.BigInteger;
import java.util.HashMap;
import java.util.Scanner;

public class D3 {
	public void solve(Scanner in, PrintWriter out) {
		int n = in.nextInt();
		long cumsum = 0;
		BigInteger ans = new BigInteger("0");
		HashMap<Long, Integer> count = new HashMap<>();
		for(int i = 1; i <= n; ++i) {
			long x = in.nextInt();
			cumsum += x;
			count.put(x, count.getOrDefault(x, 0) + 1);
			
			long curCount = i;
			long curSum = cumsum;
			
			curCount -= count.get(x);
			curCount -= count.getOrDefault(x - 1, 0);
			curCount -= count.getOrDefault(x + 1, 0);
			
			curSum -= x * count.get(x);
			curSum -= (x - 1) * count.getOrDefault(x - 1, 0);
			curSum -= (x + 1) * count.getOrDefault(x + 1, 0);
			
			ans = ans.add(BigInteger.valueOf(curCount * x - curSum));
		}
		
		out.println(ans);
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		new D3().solve(in, out);

		in.close();
		out.close();
	}
}
