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 <iostream>
using namespace std;
void swap(int*a,int*b){
  int tmp = *a;
  *a=*b;
  *b=tmp;
}
int main(void){
    int loop;
    cin >> loop;
    int cnt=0;
    int* n = new int[loop];
    for(int i=0;i<loop;i++) 
        cin >> n[i]; 
    for(int i=0;i<loop;i++) {
        for(int j=0;j<loop-i-1;j++) {
            if(n[j] > n[j+1]) {
                swap(n[j],n[j+1]);
                cnt++;
            }
        }
    }
    cout << cnt << endl;
    return 0;
}