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>
#include <stdlib.h>

long count=0;

typedef struct number{
  long num;
  struct number *next;
}Number;

Number *create(long a, Number *begin) {
  Number *p = (Number*)malloc(sizeof(Number));
  p->num = a;
  p->next = NULL;
  if(begin == NULL) {
    return p;
  }
  if(begin->num < a) {
    p->next = begin;
    return p;
  }
  count++;
  Number *p0 = begin;
  while(p0->next != NULL) {
    if(p0->next->num < a) break;
    p0=p0->next;
    count++;
  }
  p->next = p0->next;
  p0->next = p;
  return begin;
}

int main() {
  long i,n,a;
  Number *begin;
  scanf("%ld",&n);
  for(i = 0; i < n; i++) {
    scanf("%ld",&a);
    begin = create(a,begin);
  }
  printf("%ld\n",count);

  return 0;
}