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
n=int(input())
v =list(map(int,input().split()))
mp=defaultdict(lambda : 0)
bp=defaultdict(lambda : 0)
b=[0 for i in range(n)]
for i in range(n):
	l=i-mp[v[i]]-mp[v[i]+1]-mp[v[i]-1]
	b[i]+=l
	mp[v[i]]+=1
for i in range(n):
	l=i-bp[v[n-1-i]]-bp[v[n-1-i]-1]-bp[v[n-1-i]+1]
	# print(l)
	b[n-1-i]-=l
	bp[v[n-1-i]]+=1
ans=0
# print(bp)
for i in range(n):
	# print(b[i],v[i])
	ans+=b[i]*v[i]
print(ans)