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(input())
arr = [int(e) for e in input().split()]
m={}
for i in arr:
    if i not in m:
        m[i] = 0
    m[i] += 1
ans = 0
f = {}
for i in range(n):
    tot = n
    left = i
    right = n-i-1
    if arr[i] not in f:
        f[arr[i]] = 0
    if (arr[i]+1) not in f:
        f[arr[i]+1] = 0
    if (arr[i]-1) not in f:
        f[arr[i]-1] = 0
    if arr[i] not in m:
        m[arr[i]] = 0
    if (arr[i]+1) not in m:
        m[arr[i]+1] = 0
    if (arr[i]-1) not in m:
        m[arr[i]-1] = 0
    left -= f.get(arr[i])
    left -= f.get(arr[i]-1)
    left -= f.get(arr[i]+1)
    f[arr[i]] += 1
    right -= (m.get(arr[i])-f.get(arr[i]))
    right -= (m.get(arr[i]+1)-f.get(arr[i]+1))
    right -= (m.get(arr[i]-1)-f.get(arr[i]-1))
    ans += (left*arr[i] - right*arr[i])
print(ans)