Let's denote a function 

<image>

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
n=int(input())
a1=list(map(int,input().split()))
p=[0 for _ in range(n+1)]
a=[0 for _ in range(n+1)]
for i in range(1,n+1):
	a[i]=a1[i-1]
for i in range(1,n+1):
	p[i]=p[i-1]+a[i]
ans=0
for i in range(1,n):
	ans=ans+p[n]-2*p[i]
c=0
# print(ans)
d = defaultdict(int)
d[a[n]]+=1
for i in range(n-1,0,-1):
	d[a[i]]+=1
	if(d[a[i]+1]>0):
		ans-=d[a[i]+1]
	if(d[a[i]-1]>0):
		ans+=d[a[i]-1]
print(ans)