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 200000

int sum(int);
void add(int);
void solve(void);

int n, a[MAX], bit[MAX];

int main(){
  int i;

  scanf("%d" ,&n);
  for(i = 0 ; i < n ; i++){
    scanf("%d" ,&a[i]);
  }
  solve();

  return 0;
}

int sum(int i){
  int s = 0;

  while(i > 0){
    s += bit[i];
    i -= i & -i;
  }
  return s;
}

void add(int i){
  while(i <= n){
    bit[i]++;
    i += i & -i;
  }
}

void solve(){
  int i, ans = 0;

  for(i = 0 ; i < n ; i++){
    ans += i - sum(a[i]);
    add(a[i]);
  }
  printf("%d\n" ,ans);
}