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. 
n=int(raw_input())
N=map(int,raw_input().split())
d={}
for i in N:
    d[i]=0
for i in N:
    d[i]+=1
s=[N[0]]
for i in range(1,n):
    s.append(N[i]+s[i-1])
su=sum(N)
ans=0
for i in range(n):
    foo=d[N[i]]*(N[i])
    x=d[N[i]]
    d[N[i]]-=1
    if N[i]+1 in d:
        foo+=d[N[i]+1]*(N[i]+1)
        x+=d[N[i]+1]
    if N[i]-1 in d:
        foo+=d[N[i]-1]*(N[i]-1)
        x+=d[N[i]-1]
    #print "no of times %d, value %d , index %d, sum %d" %(x,foo,i,su)
    ans+=(su-foo-(n-x-i)*N[i])
    #print "ans %d" %(ans)
    su-=N[i]
print ans

