Problem:
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The Number of Inversions

For a given sequence $A = \{a_0, a_1, ... a_{n-1}\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:

bubbleSort(A)
cnt = 0 // the number of inversions
for i = 0 to A.length-1
for j = A.length-1 downto i+1
if A[j] < A[j-1]
swap(A[j], A[j-1])
cnt++

return cnt

For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.

Input

In the first line, an integer $n$, the number of elements in $A$, is given. In the second line, the elements $a_i$ ($i = 0, 1, .. n-1$) are given separated by space characters.

output

Print the number of inversions in a line.

Constraints

$ 1 \leq n \leq 200,000$

$ 0 \leq a_i \leq 10^9$

$a_i$ are all different

Sample Input 1

5
3 5 2 1 4

Sample Output 1

6

Sample Input 2

3
3 1 2

Sample Output 2

2

Solution in Ruby (judge verdict: Accepted)

$invNum =0
def merge(left, mid, right)
    leftArr = $arr[left..mid]
    rightArr = $arr[mid+1..right]
    leftArr << 1000000001
    rightArr << 1000000001
    l = r = 0
    (left..right).each {|i|
        if leftArr[l] <= rightArr[r]
            $arr[i] = leftArr[l]
            l += 1
        else
            $arr[i] = rightArr[r]
            r += 1
            $invNum += mid - left - l + 1
        end
    }
end

def mergeSort(left, right)
    mid = (left + right)/2
    if left + 1 < right
        mergeSort(left, mid)
        mergeSort(mid+1, right)
        merge(left, mid, right)
    elsif left + 1 == right
        merge(left, mid, right)
    end
end

N = gets.to_i
$arr = gets.split.map &:to_i
mergeSort(0,N-1)
puts $invNum