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 collections import defaultdict

freq = defaultdict(int)

n = int(raw_input())

a = map(int, raw_input().split())

for i in a:
    freq[i] += 1

suffix_sum = [0]*n
suffix_sum[n-1] = a[-1]

for i in xrange(n-2, -1, -1):
    suffix_sum[i] = suffix_sum[i+1] + a[i]

ans = 0
#print suffix_sum
for i in xrange(n-1):
    temp1 = freq[a[i]+1]
    temp2 = freq[a[i]-1]
    temp3 = suffix_sum[i+1]
    #print temp1, temp2, ans
    ans = ans + temp3 - (n-1-i)*a[i] - max(temp1, 0) + max(temp2, 0)
    freq[a[i]] -= 1

print ans
