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. 
m = {}
A = [0] * 200001	
B = []
sum1 = 0
ans = 0
n = int(input())
B = (input().split(' '))
for i in range(n):
	A[i] = int(B[i])
	sum1 += A[i]
	if A[i] in m:
		m[A[i]] += 1
	else:
		m[A[i]] = 1
np = n
for i in range(n):
	if A[i] in m:
		a1 = m[A[i]]
	else:
		a1 = 0
	if A[i] + 1 in m:
		a2 = m[A[i] + 1]
	else:
		a2 = 0
	if A[i] - 1 in m:
		a3 = m[A[i] - 1]
	else:
		a3 = 0
	val1 = (sum1 - A[i] * a1 - (A[i] - 1) * a3 - (A[i] + 1) * a2)
	val2 = (np - a1 - a3 - a2) * A[i]
	ans += (val1 - val2)
	# print((sum1 - A[i] * a1 - (A[i] - 1) * a3 - (A[i] + 1) * a2) - (np - a1 - a3 - a2) * A[i]
	m[A[i]] -= 1;
	sum1 -= A[i];
	np -= 1;
print(ans)