Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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
import Data.List (foldl')

main = do
    n <- readLn
    xs <- map read.words<$>getLine :: IO [Int]
    print $ foldl' (\n x -> n + snd x) 0 . mSort n . zip xs $ repeat ( 0 :: Int)

mergeSort xs = mSort n xs
    where n = length xs

mSort :: (Ord a) => Int -> [(a,Int)] -> [(a,Int)]
mSort 0 [] = []
mSort 1 [x] = [x]
mSort n xs = merge nl nr sortedLeft sortedRight
    where nl = n `div` 2
          nr = n - nl
          (left, right) = splitAt nl xs
          sortedLeft = mSort nl left
          sortedRight = mSort nr right

merge ll 0 xs ys = xs
merge 0 lr [] ys = ys
merge ll lr (x:xs) (y:ys)
  | fst x < fst y = x : merge (ll-1) lr xs (y:ys)
  | otherwise     = (fst y, snd y + ll) : merge ll (lr - 1) (x:xs) ys