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>
#include <string.h>
#include <ctype.h>
#include <math.h>

//============================================================================
#define pred(x) ((x)-1)
#define succ(x) ((x)+1)
//============================================================================
//============================================================================
//============================================================================
void swap(long *a, long *b)
{
    long c;

    c=*a, *a=*b, *b=c;
}
//----------------------------------------------------------------------------
long bubbleSort(long *A, long N)
{
    long cnt,i,j,k,l,S;

    /*for (cnt=i=0, l=N; i<pred(N); i++, l--) // ?????¢??°
    for (j = pred(l); i<j; j--)
        if (A[j] < A[pred(j)]){
	        swap(&A[j], &A[j-1]);
	        cnt++;
        }*/
    for (i=cnt=0; i<pred(N); i++)
    for (j=succ(i); j<N; j++){
        if (A[j]<A[i]){
            cnt++;
        }
	}
    return cnt;
}
//----------------------------------------------------------------------------
long bubblesort(long *a, long N)
{
	long i, j, k;
	long cnt,x;

	k = N - 1;  cnt=0;
	while (k >= 0) {
		j = -1;
		for (i = 1; i <= k; i++)
			if (a[i - 1] > a[i]) {
				j = i - 1;
				x = a[j];  a[j] = a[i];  a[i] = x;
                cnt++;
			}
		k = j;
	}
    return cnt;
}
//----------------------------------------------------------------------------
long cnt;
void quicksort(long *a, long first, long last)
{
	long i, j;
	long x, t;

	x = a[(first + last) / 2];
	i = first;  j = last;
	for ( ; ; ) {
		while (a[i] < x) i++;
		while (x < a[j]) j--;
		if (i >= j) break;
		t = a[i];  a[i] = a[j];  a[j] = t; cnt++;
		i++;  j--;
	}
	if (first  < i - 1) quicksort(a, first , i - 1);
	if (j + 1 < last) quicksort(a, j + 1, last);
}
//----------------------------------------------------------------------------
void input(FILE *file)
{
    long N,n,*ary;

    fscanf(file, "%ld", &N);
    ary = malloc(succ(N)*sizeof(long));
	for (n = 0; n < N; n++) {
        fscanf(file, "%ld", &ary[n]);
	}
    //quicksort(ary, 0, N-1);
    printf("%ld\n", bubblesort(ary, N));
    free(ary);
}
//============================================================================
#ifndef DESKTOP
int main()
{
    input(stdin);
    return 0;
}
#endif