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, a, ft = 0, [(0, 0)], [0]

def query(k):
    res = 0
    while k > 0:
        res += ft[k]
        k -= k&-k
    return res
    
def adjust(k, v):
    while k <= n:
        ft[k] += v
        k += k&-k
        
def rquery(l, r):
    if l > r:
        return 0
    return query(r) - (0 if l == 1 else query(l-1))


n, ans = int(input()), 0
tmp = [int(x) for x in input().split()]
for i in range(1, n+1):
    a.append((tmp[i-1], i))
ft = [0 for i in range(n+1)]
ans = sum([x*(2*i-n-1) for x, i in a])
a.sort(key=lambda x: x[0])

L, R, cur, i = [], [], 0, 1
while i <= n:
    v, k = a[i]
    if v == cur:
        ans -= rquery(1, k-1) - rquery(k+1, n)
        R.append(k)
        i += 1
    else:
        for j in L:
            adjust(j, -1)
        L = []
        if v - cur == 1:
            for j in R:
                L.append(j)
                adjust(j, 1);
        R, cur = [], v
print(ans)
    