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 CF930D{

	static int mx = 212345;
	static int[] v = new int[mx];
	static TreeMap<Integer, Integer> cnt = new TreeMap<>();
	
	static int count(int key){
		if(!cnt.containsKey(key)) return 0;
		return cnt.get(key);
	}
	public static void main (String[] args)
	{
		Scanner sc = new Scanner(System.in);
		int n;
		n = sc.nextInt();
		for (int i = 0; i < n; i++) {
			v[n-1-i] = sc.nextInt();
		}
		BigInteger ans = BigInteger.ZERO;
		long x = 0, y, z;
		int a, b, c;
		for (int i = 0; i < n; i++) {
			a = count(v[i]-1);
			b = count(v[i]);
			c = count(v[i]+1);
			cnt.put(v[i], ++b);
			x += v[i];
			y = (long)a*(v[i]-1) + (long)b*(v[i]) + (long)c*(v[i]+1);
			z = ((i+1)-(a+b+c))*(long)v[i];
			ans = ans.add(BigInteger.valueOf(x));
			ans = ans.subtract(BigInteger.valueOf(y));
			ans = ans.subtract(BigInteger.valueOf(z));
		}
		System.out.println(ans);
	}
}