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()
a=input()
n=int(n)
import collections
l=list(map(int,a.split(' ')))
for i in range(0,int(n)-1):
    l[i]=int(l[i])
cnt=collections.Counter()
cnt[l[0]]+=1;
b=list()
b.append(l[0])
ans=0
for i in range(1,n):
    b.append(int(b[i-1]+l[i]))
for i in range(1,n):
    v1=cnt[l[i]]
    v2=cnt[l[i]+1]
    v3=cnt[l[i]-1]
    ans=ans+((i-v1-v2-v3)*l[i]-(b[i-1]-v1*l[i]-v2*(l[i]+1)-v3*(l[i]-1)))
    cnt[l[i]]+=1
print(ans)
