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>
const int MAX_N = 200000;

int main(void) {
  int n, cnt = 0;
  int i, j;
  int ary[MAX_N];

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

  for (i = 0; i < n; i++) {
    for (j = n-1; j > i; j--) {
      if (ary[j] < ary[j-1]) {
        int tmp = ary[j];
        ary[j] = ary[j-1];
        ary[j-1] = tmp;
        cnt++;
      }
    }
  }

  printf("%d\n", cnt);

  return 0;
}