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.*;
import java.math.*;
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    while (sc.hasNext()) {
      int n = sc.nextInt();
      int[] a = new int[n];
      HashMap<Integer, Integer> cnt = new HashMap<>();
      BigInteger ans = BigInteger.ZERO;
      for (int i=0; i<n; ++i) {
        a[i] = sc.nextInt();
        ans = ans.add(
            new BigInteger(""+((long)a[i]*(2*i+1-n))));
        if (!cnt.containsKey(a[i])) cnt.put(a[i], 0);
        cnt.put(a[i], 1+cnt.get(a[i]));
      }
      for (int i=0; i<n; ++i) {
        if (cnt.containsKey(a[i]-1))
          ans = ans.add(
              new BigInteger(""+cnt.get(a[i]-1)));
        if (cnt.containsKey(a[i]+1))
          ans = ans.subtract(
              new BigInteger(""+cnt.get(a[i]+1)));
        cnt.put(a[i], cnt.get(a[i])-1);
      }
      System.out.println(ans.toString());
    }
  }
}
