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.util.Scanner;
import java.math.BigInteger;
import java.util.*;
import java.util.Scanner;

public class Main {
	public static void main (String[] args) {
		Scanner rd = new Scanner(System.in);
		HashMap<Integer,Integer> hs = new HashMap<>();
		BigInteger sum,ans;
		sum = BigInteger.valueOf(0);
		ans = BigInteger.valueOf(0);
		int n = rd.nextInt();
	    int[] arr = new int[2000050];
	    for(int i=0;i<n;i++)
	    {
	        arr[i] = rd.nextInt();
	        ans = ans.add((BigInteger.valueOf(arr[i]).multiply(BigInteger.valueOf(i))));
	        ans = ans.subtract(sum);
	        if(hs.containsKey(Integer.valueOf(arr[i]+1)))
	            ans = ans.add(BigInteger.valueOf(hs.get(Integer.valueOf(arr[i]+1))));
	        if(hs.containsKey(Integer.valueOf(arr[i]-1)))
	        ans = ans.subtract(BigInteger.valueOf(hs.get(Integer.valueOf(arr[i]-1))));
	        sum = sum.add(BigInteger.valueOf(arr[i]));
	        if(hs.containsKey(Integer.valueOf(arr[i])))
	        {
	            Integer val = hs.get(Integer.valueOf(arr[i]));
	            val++;
	            hs.put(Integer.valueOf(arr[i]),val);
	        }
	        else
	        {
	            hs.put(Integer.valueOf(arr[i]),Integer.valueOf(1));
	        }
	        
	    }
	    System.out.println(ans);
	}
}