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 = input()
arr = []
arr = map(int, raw_input().split())
mp = {}
sum = 0
ans = 0
for i in range(n):
	i1 = i + 1
	sum += arr[i]
	curr = arr[i]
	if curr in mp > 0:
		mp[curr] = mp[curr] + 1
	else:
		mp[curr] = 1
	c1 = 0
	c2 = 0
	c3 = 0
	if curr in mp:
		c1 = mp[curr]
	if curr - 1 in mp:
		c2 = mp[curr - 1]
	if curr + 1 in mp:
		c3 = mp[curr + 1]
	num = i1 - (c1 + c2 + c3)
	val = num * curr
	c1 = c1 * curr
	c2 = c2 * (curr - 1)
	c3 = c3 * (curr + 1)
	diff = sum - (c1 + c2 + c3)
	diff1 = val - diff
	ans = ans + diff1
print ans