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)

// AOJ ALDS1_5_D The Number of Inversions
// 2018.4.30 bal4u

#include <stdio.h>
#include <string.h>

#if 1
#define gc() getchar_unlocked()
#else
#define gc() getchar()
#endif
int in()
{
	int n = 0, c = gc();
	do n = 10*n + (c & 0xf), c = gc(); while (c >= '0');
	return n;
}

#define MAX_N 200002
#define INF   0x7f7f7f7f

int a[MAX_N];
int w[MAX_N/2+2];

long long number_of_inversions(int low, int high) 
{ 
	int i, j, k, p, mid;
	long long ans = 0;

	if (low < high) {
		mid = (low + high) >> 1;
		ans += number_of_inversions(low,   mid);
		ans += number_of_inversions(mid+1, high);

		p = mid-low+1;
		memcpy(w, a+low, p << 2);
		i = mid+1, j = 0; k = low;

		w[p] = INF;
		while (i <= high) {
			if (w[j] <= a[i]) a[k++] = w[j++];
			else              a[k++] = a[i++], ans += i-k;
		}
		memcpy(a+k, w+j, (p-j) << 2);
	}
	return ans;
}

int main()
{
	int n, i;
	
	n = in(); for (i = 0; i < n; i++) a[i] = in();
	printf("%lld\n", number_of_inversions(0, n-1));
	return 0;
}