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 Python3 (judge verdict: Accepted)

def g(A,l,m,r):
 global c
 L,R=A[l:m]+[1e9],A[m:r]+[1e9]
 i=j=0
 for k in range(l,r):
  if L[i]<R[j]:A[k]=L[i];i+=1
  else:A[k]=R[j];j+=1;c+=m-l-i
 return c
def s(A,l,r):
 if l+1<r:
  m=(l+r)//2
  return s(A,l,m)+s(A,m,r)+g(A,l,m,r)
 else:return 0
c=0
n=int(input())
A=list(map(int,input().split()))
s(A,0,n)
print(c)