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())
a = [0] + list(map(int, input().split()))
s = [0] * len(a)
mp = {}
for i in range(1, n + 1):
    s[i] = s[i - 1] + a[i]
    if mp.get(a[i]):
        mp[a[i]] += 1
    else:
        mp[a[i]] = 1
ans = 0
for i in range(1, n + 1):
    ans += s[n] - s[i - 1]
    l = mp.get(a[i] - 1, 0)
    m = mp.get(a[i], 0)
    r = mp.get(a[i] + 1, 0)
    ans -= l * (a[i] - 1) + m * a[i] + r * (a[i] + 1)
    ans -= a[i] * (n - i + 1 - l - m - r)
    mp[a[i]] -= 1
    if mp[a[i]] == 0:
        mp.pop(a[i])
print(ans)
    
