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. 
inp = []
d = {}
n = (int)(input())
tot=0
ans=0
temp = input()
temp = temp.split()
inp = [(int)(i) for i in temp]
for i in range(n):
	val = inp[i]
	tot = tot+val
	if val not in d: d[val] = 1
	else: d[val] += 1
for i in range(n-1,0,-1):
	prev = 0
	nxt = 0
	if inp[i]+1 in d: nxt = d[inp[i]+1]
	if inp[i]-1 in d: prev = d[inp[i]-1]
	ans += inp[i]*(i+1 - d[inp[i]] - prev - nxt)
	ans -= tot - inp[i]*d[inp[i]] - (inp[i]+1)*nxt - (inp[i]-1)*prev
	d[inp[i]] -= 1
	tot -= inp[i]
print(ans)
