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.util.*;
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());
        st = new StringTokenizer(br.readLine());
        HashMap<Long,Long> h = new HashMap<Long,Long>();
        long[] a = new long[n];
        for(int i=0;i<n;i++){
            a[i] = Long.parseLong(st.nextToken());
            long k = 1;
            if(h.containsKey(a[i])) k += h.get(a[i]);
            h.put(a[i],k);
        }
        long[] s = new long[n];
        s[0] = 0;
        for(int i=1;i<n;i++) s[i] = s[i-1]+a[i-1];
        long sum;
        long nb;
        long rep = 0;
        BigInteger b = new BigInteger("0");
        BigInteger bp = new BigInteger("0");
        for(int i=n-1;i>0;i--){
            sum = s[i];
            nb = h.get(a[i]);
            nb--;
            if(nb==0) h.remove(a[i]);
            else h.put(a[i],nb);
            Long adad = (long) i;
            adad -= nb;
            sum -= nb*a[i];
            nb = 0;
            if(h.containsKey(a[i]-1)) nb = h.get(a[i]-1);
            sum -= nb*(a[i]-1);
            adad -= nb;
            nb = 0;
            if(h.containsKey(a[i]+1)) nb = h.get(a[i]+1);
            sum -= nb*(a[i]+1);
            adad -= nb;
            rep = adad*a[i]-sum;
            String ss = new String();
            ss = ss.valueOf(rep);
            bp = new BigInteger(ss);
            b = b.add(bp);
        }
        System.out.println(b.toString());
    }
}