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)

using System;

class Program{
    static void Merge(int[] A, int left, int mid, int right){
        var n1=mid-left;
        var n2=right-mid;
        var L=new int[n1+1];
        var R=new int[n2+1];
        for(var i=0; i<n1; i++){
            L[i]=A[left+i];
        }
        for(var i=0; i<n2; i++){
            R[i]=A[mid+i];
        }
        L[n1]=1000*1000*1000+1;
        R[n2]=1000*1000*1000+1;
        var l=0;
        var j=0;
        for(var k=left; k<right; k++){
            if(L[l]<=R[j]){
                A[k]=L[l];
                l++;
            } else{
                A[k]=R[j];
                j++;
            }
        }
    }

    static int BinarySearch(int[] A, int val, int left, int right){
        if(val>A[right-1]){
            return right;
        }
        while(left<=right){
            var mid=(left+right)/2;
            if(A[mid]<val){
                left=mid+1;
            } else{
                right=mid-1;
            }
        }
        return left;
    }

    static long MergeSort(int[] A, int left, int right){
        if(left+1<right){
            var mid=(left+right)/2;
            var a=MergeSort(A, left, mid);
            var b=MergeSort(A, mid, right);
            long c=0;
            for(var i=left; i<mid; i++){
                c+=BinarySearch(A, A[i], mid, right)-mid;
            }
            Merge(A, left, mid, right);
            return a+b+c;
        } else{
            return 0;
        }
    }

    public static void Main(){
        var n=int.Parse(Console.ReadLine());
        var s=Console.ReadLine().Split(' ');
        var S=new int[n];
        for(var i=0; i<n; i++){
            S[i]=int.Parse(s[i]);
        }

        var result=MergeSort(S, 0, n);

        Console.WriteLine(result);
    }
}