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: Wrong Answer)

#include <stdio.h>

long int cnt = 0;

void merge(int a[], int left, int mid, int right) {
  int n1 = mid - left;
  int n2 = right - mid;
  int L[2000000];
  int R[2000000];
  for (int i = 0; i < n1; i++) {
    L[i] = a[left + i];
  }
  for (int i = 0; i < n2; i++) {
    R[i] = a[mid + i];
  }
  L[n1] = 10000000001;
  R[n2] = 10000000001;

  int i = 0;
  int j = 0;
  for (int k = left; k < right; k++) {
    if (L[i] <= R[j]) {
      a[k] = L[i++];
    } else {
      a[k] = R[j++];
      cnt += n1 - i;
    }
  }
}

void mergeSort(int a[], int left, int right) {
  int mid = 0;
  mid = (left + right) / 2;
  if (left + 1 < right) {
    mergeSort(a, left, mid);
    mergeSort(a, mid, right);
    merge(a, left, mid, right);
  }
}

int main() {
  int n;
  scanf("%d\n", &n);
  int a[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  mergeSort(a, 0, n);
  printf("%d\n", cnt);

  return 0;
}