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 cf903D {
	static int getMap (HashMap<Long, Integer> mp, long x) {
		Integer val = mp.get(x);
		if (val == null) {
			return 0;
		}
		return val;
	}

	public static void main (String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		HashMap<Long, Integer> cnt = new HashMap<Long, Integer> ();

		BigInteger ans = BigInteger.ZERO;
		BigInteger sum = BigInteger.ZERO;

		for (int i = 0; i < n; i++) {
			long x = Long.parseLong(st.nextToken());
			ans = ans.add(BigInteger.valueOf(i * x - getMap(cnt, x - 1) + getMap(cnt, x + 1))).subtract(sum);
			sum = sum.add(BigInteger.valueOf(x));
			int ncnt = getMap(cnt, x) + 1;
			cnt.put(x, ncnt);
		}

		System.out.println(ans);
	}
}