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 Haskell (judge verdict: Time Limit Exceeded)

import Control.Applicative
import Control.Monad.State

main = do
    n <- getLine
    xs <- map read.words<$>getLine
    let (list, s) = bubbleSort xs
    print s

bubbleSort xs = runState (bSort xs) 0

bSort :: [Int] -> State Int [Int]
bSort [] = return []
bSort xs = do
    (y:ys) <- bubble xs
    ys' <- bSort ys
    return (y:ys')

bubble :: [Int] -> State Int [Int]
bubble [x] = return [x]
bubble (x:xs) = do
    (y:ys) <- bubble xs
    if x > y
       then modify (+1) >> return (y:x:ys)
       else return (x:y:ys)