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=int(raw_input())
arr=[]
x=raw_input().split(' ')
i=0
while i<n:
    arr.append(int(x[i]))
    i=i+1

mp={}
totSum=0
sumi=0;

i=n-1
while i>=0:
    after=arr[i]+1;
    before=arr[i]-1;
    eq=arr[i];
    cntA=0;
    cntB=0;
    cntE=0;
    
    if (after in mp):
        cntA=mp[after]

    if (before in mp):
        cntB=mp[before]

    if (eq in mp):
        cntE=mp[eq]


    v=sumi-((cntA*after) + (cntB*before) + (cntE*eq));
    ctr=(n-1-i)-(cntA+cntB+cntE);
    totSum=totSum + (v-(ctr*arr[i]));
    sumi=sumi+arr[i]
    
    mp[arr[i]]=cntE+1
    
    i=i-1
    
print totSum
