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. 
from collections import defaultdict

n=int(input())
m=input().split()
v=[int(x) for x in m]
pre = [0]*n
temp = [0]*n
mp = defaultdict(lambda : 0)
ans = 0
L = 0
R = 0
curr = 0
for x in v:
    mp[x] += 1

for i in range (n-1):
    pre[i+1] += 1
    temp[i+1] -= v[i]
    
for i in range(1,n):
    pre[i] += pre[i-1]
    temp[i] += temp[i-1]
for i in range(n-1, 0, -1):
        mp[v[i]] -= 1
        L = mp[v[i]-1]
        R = mp[v[i]+1]
        curr = mp[v[i]]
        ans += temp[i]
        ans += ((pre[i] - (L + R + curr)) * v[i])
        ans += L * (v[i] - 1) + R * (v[i] + 1) + curr * v[i]
    

    
print(ans)