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>
#include <stdlib.h>
#define N 200010
int n;
int a[N];
int cnt = 0;
void merge(int a[], int left, int mid, int right){
    int i = left;
    int j = mid+1;
    int k = 0;
    int t[n];
    while(i <= mid && j <= right){
        if(a[i] <= a[j]) t[k++] = a[i++];
        else{
            cnt += mid - i + 1;
            t[k++] = a[j++];
        }
    }
    while(i <= mid) t[k++] = a[i++];
    while(j <= right) t[k++] = a[j++];

    for(int p = 0; p < k; p++){
        a[left + p] = t[p];
    }
}
void mergeSort(int a[], int left, int right){
    if(left < right){
        int mid = (left + right) / 2;
        mergeSort(a, left, mid);
        mergeSort(a, mid+1, right);
        merge(a, left, mid, right);
    }
}
int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }

    mergeSort(a, 0, n-1);
    printf("%d\n", cnt);
//    for(int i = 0; i < n; i++) {
//        printf("%d ", a[i]);
//    }
    return 0;
}