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>

long merge(int a[],int left,int mid,int right){
    int n1,n2,i,j,k;
    int l[250002],r[250002];
    long cccc = 0;
    n1 = mid-left; n2 = right-mid;
    for(i=0;i<n1;i++){
        l[i] =a[i+left];
    }
    for(i=0;i<n2;i++){
        r[i] =a[i+mid];
    }
    l[n1]=1000000001;r[n2]=1000000001;
    i=0;j=0;
    for(k=left;k<right;k++){
        if(l[i]<=r[j]){
            a[k] = l[i];
            i++;
        }else{
            a[k] = r[j];
            j++;
            cccc+=n1-i;
        }
    }
    return cccc;
}

long mergesort(int a[],int left,int right){
    int mid;
    long c1=0,c2=0,c3=0;
    if(left+1<right){
        mid = (left+right)/2;
        c1=mergesort(a,left,mid);
        c2=mergesort(a,mid,right);
        c3=merge(a,left,mid,right);
    }
    return c1+c2+c3;
}

int main(void){
    int n,i;
    int a[200000];
    long cccc=0;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    cccc = mergesort(a,0,n);
    printf("%ld\n",cccc);
}