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 C (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#define MAX_N 200000
int a[MAX_N];
int bit[MAX_N+1],n;
int sum(int i){
  int s=0;
  while(i>0){
    s+=bit[i];
    i-=i & -i;
  }
  return s;
}
void add(int i,int x){
  while(i<=n){
    bit[i]+=x;
    i+=i & -i;
  }
}
int main(){
  long long int ans=0;
  int i,j;
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%d",&a[i]);
  for(j=0;j<n;j++){
    ans+=j-sum(a[j]);
    add(a[j],1);
  }
  printf("%d\n",ans);
  return 0;
}