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 get_uint() {
    int n = 0;
    int c = getchar_unlocked();
    if(c < 48 || 57 < c) return c;
    while(47 < c && c < 58) {
        n = 10 * n + (c & 0xf);
        c = getchar_unlocked();
    }
    return n;
}
void put_ull(long long n) {
    if(!n) {
        putchar_unlocked('0');
        return;
    }
    char buf[20];
    int i = 0;
    while(n) buf[i++] = (char)(n % 10 + '0'), n /= 10;
    while(i--)putchar_unlocked(buf[i]);
}
long long merge_sort(int *first, int *last) {
    const int N = (int)(last - first);
    if(N < 2) return 0;
    int *mid = first + N / 2;
    const int N1 = (int)(mid - first);
    const int N2 = (int)(last - mid);
    int *left = malloc(sizeof(int) * (N1 + 1));
    int *right = malloc(sizeof(int) * (N2 + 1));
    memcpy(left, first, sizeof(int) * (N1 + 1));
    memcpy(right, mid, sizeof(int) * (N2 + 1));
    left[N1] = right[N2] = 1 << 30;

    long long cnt = 0;
    cnt += merge_sort(left, left + N1);
    cnt += merge_sort(right, right + N2);

    for(int *it = first, i = 0, j = 0; it != last; ++it) {
        if(left[i] <= right[j]) *it = left[i++];
        else *it = right[j++], cnt += N / 2 - i;
    }
    free(left);
    free(right);
    return cnt;
}
int main(int argc, char **argv) {
    int n = get_uint();
    int a[200001];
    for(int i = 0; i < n; ++i) a[i] = get_uint();
    put_ull(merge_sort(a, a + n));
    putchar_unlocked('\n');
    return 0;
}