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

public class te {
	public static void main(String[] args){
		int []a = new int [201000];
		long []sum = new long [201000];
		BigInteger ans = BigInteger.ZERO;
		Scanner cin=new Scanner(System.in);
        Map<Integer,Integer> mp=new HashMap<Integer,Integer>();
		int n = cin.nextInt();
		for(int i=0;i<n;++i)
			a[i] = cin.nextInt();
		for(int i=n-1;i>=0;--i){
			sum[i] = sum[i+1]+a[i];
			//System.out.println("a[i] == " + a[i] + " sum[i+1] = " + sum[i+1]);
			ans = ans.add(BigInteger.valueOf(sum[i+1]));
			ans = ans.subtract(BigInteger.valueOf(a[i]).multiply(BigInteger.valueOf(n-1-i)));
			if(mp.containsKey(a[i]-1)) ans = ans.add(BigInteger.valueOf(mp.get(a[i]-1)));
			if(mp.containsKey(a[i]+1)) ans = ans.subtract(BigInteger.valueOf(mp.get(a[i]+1)));
			if(mp.containsKey(a[i])) mp.put(a[i], mp.get(a[i])+1);
			else mp.put(a[i], 1);
		}
		System.out.println(ans);
	}
}