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. 
//package codeforces;
import java.math.*;
import java.util.*;
public class Main {
	static int n;
	static long []a=new long[200005];
	public static void main(String args[]){
		Scanner cin=new Scanner(System.in);
		while(cin.hasNext()){
			n=cin.nextInt();
			Map<Long,Integer> map=new HashMap();
			BigInteger ans=BigInteger.valueOf(0);
			for(int i=1;i<=n;i++){
				a[i]=cin.nextLong();
				if(map.containsKey(a[i]-1)){
					ans=ans.subtract(BigInteger.valueOf(map.get(a[i]-1)));
				}
				if(map.containsKey(a[i]+1)){
					ans=ans.add(BigInteger.valueOf(map.get(a[i]+1)));
				}
				if(map.containsKey(a[i])){
					map.put(a[i],map.get(a[i])+1);
				} else{
					map.put(a[i],1);
				}
			}
			//System.out.println(ans);
			BigInteger tt=BigInteger.valueOf(0);
			for(int i=2;i<=n;i++){
				tt=tt.add(BigInteger.valueOf((i-1)*(a[i]-a[i-1])));
				ans=ans.add(tt);
			}
			System.out.println(ans);
		}
		cin.close();
	}
}
