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: Accepted)

#include <stdio.h>
#define SENTINEL 2000000000
long long  a[200010],count = 0;
void merge(int left,int mid,int right)
{
    int n1 = mid - left, n2 = right - mid;
    int l[n1 + 10],r[n2 + 10];
    int i,j,k;
    for(i = 0;i < n1;i++)
        l[i] = a[left + i];
    for(i = 0;i < n2;i++)
        r[i] = a[mid + i];
    i = j = 0;
    l[n1] = r[n2] = SENTINEL;
    for(k = left;k < right;k++)
    {
        if(l[i] < r[j])
            a[k] = l[i++];
        else
        {
            count+=n1 - i;
            a[k] = r[j++];
        }
    }
}
void mergesort(int left,int right)
{
    if(left + 1 < right)
    {
        long long mid = (left + right) / 2;
        mergesort(left,mid);
        mergesort(mid,right);
        merge(left,mid,right);
    }
}
int main()
{
    int n,i;
    scanf("%d",&n);
    for(i = 0;i < n;i++)
        scanf("%lld",&a[i]);
    mergesort(0,n);
    printf("%lld\n",count);
    return 0;
}