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. 
from random import randint
from time import time
n = list(map(int, input().strip().split()))[0]
arr = list(map(int, input().strip().split()))

tik = time()
count = dict()
subsum = [0 for i in range(n + 1)]
for index in range(len(arr) - 1, -1, -1):
    if arr[index] in count:
        count[arr[index]] += 1
    else:
        count[arr[index]] = 1
    if index == len(arr) - 1:
        subsum[index] = arr[index]
    else:
        subsum[index] = arr[index] + subsum[index + 1]

answer = 0
for index, value in enumerate(arr):
    answer += subsum[index + 1] - ((len(arr) - index - 1) * arr[index])
    if value + 1 in count:
        answer -= count[value + 1]
    if value - 1 in count:
        answer += count[value - 1]
    count[value] -= 1
    if count[value] == 0:
        count.pop(value)
print(answer)