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. 
n = int(input())
a = input().split()

an = 0

sum = int(a[n - 1])
cnt = {}

for x in range(n):
  cnt[int(a[x])] = 0
  cnt[int(a[x]) + 1] = 0
  cnt[int(a[x]) - 1] = 0

cnt[int(a[n - 1])] = 1
x = n - 2

while x >= 0:

  rsum = sum - (cnt[int(a[x])] * int(a[x]) + cnt[int(a[x]) + 1] * (int(a[x]) + 1) + (cnt[int(a[x]) - 1] * (int(a[x]) - 1)))
  an += rsum - int(a[x]) * (n - (x + 1) - cnt[int(a[x])] - cnt[int(a[x]) + 1] - cnt[int(a[x]) - 1])

  cnt[int(a[x])] += 1
  sum += int(a[x])

  x = x - 1

print(an)