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>
#include <stdlib.h>
#include <memory.h>
 
/*
int bubbleSort(int A[], int n) {
    int i, j, t, s;
    int cnt = 0; // 反転数
    for(i=0;i<n-1;i++) {
        for(j=n-1,s=i+1;j>=s;j--) {
            if(A[j] < A[j-1]) {
                t = A[j];
                A[j] = A[j-1];
                A[j-1] = A[j];
                cnt++;
            }
        }
    }
   
    return cnt;
}
*/
 
long long int A[200005];
long long int merge_count(int l, int r){
    long long int count = 0;
    int i, j, k;
    int n = r - l;
    if(n > 1){
        long long int b[n+5];
        int m = l + n/2;
        count += merge_count(l, m);
        count += merge_count(m, r);
        memcpy((void*)b, (void*)&A[l], sizeof(b));
        for(i = l, j = 0, k = n/2; i < r; ++i){
            if(j == n/2) {
                A[i] = b[k++];
            } else if(k == n) {
                A[i] = b[j++];
            } else if(b[j] <= b[k]) {
                A[i] = b[j++];
            } else {
                A[i] = b[k++];
                count += n/2 - j;
            }
        }
    }
    return count;
}
 
long long int bubble_sort(int n) {
    return merge_count(0, n);
}
 
int main(void) {
    int i, n;
    scanf("%d\n", &n);
     
    for(i=0;i<n;i++) {
        scanf("%lld", &A[i]);
    }
     
    printf("%lld\n", bubble_sort(n));
 
    return 0;
}