Let's denote a function 

<image>

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.*;
import java.io.*;
import java.math.*;
public class Solution{

    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        PrintWriter out = new PrintWriter(System.out);
        st = new StringTokenizer(br.readLine());
        long[] a = new long[n];
        for(int i=0;i<n;i++) a[i] = Long.parseLong(st.nextToken());
        
        if(n==1) out.println("0");
        else{
            long ln = n,l;
            long rep = 0;
            BigInteger b = new BigInteger("0");
            for(int i=1;i<n;i++){
                l = i;
                String s1 = "";
                s1 = s1.valueOf((a[i]-a[i-1]));
                String s2 = "";
                s2 = s2.valueOf((ln-l)*l);
                BigInteger r = new BigInteger("1");
                r = r.multiply(new BigInteger(s1));
                r = r.multiply(new BigInteger(s2));
                b = b.add(r);
                
            }
            HashMap<Long,Integer> h = new HashMap<Long,Integer>();
            for(int i=0;i<n;i++){
                
                if(h.containsKey(a[i])){
                    int nb = h.get(a[i]);
                    nb++;
                    h.put(a[i],nb);
                }else h.put(a[i],1);
                long plus = 0;
                if(h.containsKey(a[i]+1)) plus += (long) h.get(a[i]+1);
                long moins = 0;
                if(h.containsKey(a[i]-1)) moins += (long) h.get(a[i]-1);
                rep += plus;
                rep -= moins;
                
                
            }
            
            String s = "";
            s = s.valueOf(rep);
            b = b.add(new BigInteger(s));
            
            out.println(b);
        }
        
        out.flush();
    }
    
}