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<bits/stdc++.h>
using namespace std;
#define int long long
const int MAX_N = 1 << 18;
struct BIT{
  int bit[MAX_N+1],n;
  //1-indexed
  BIT(){init();}
  BIT(int n):n(n){init();}
  void init(){
    memset(bit,0,sizeof(bit));
  }
  int sum(int i){
    int s=0;
    while(i>0){
      s+=bit[i];
      i-=i&-i;
    }
    return s;
  }
  void add(int i,int x){
    while(i<=n){
      bit[i]+=x;
      i+=i&-i;
    }
  }
  int sum0(int i){
    return sum(i+1);
  }
  void add0(int i,int x){
    add(i+1,x);
  }
};
signed main(){
  int n;
  cin>>n;
  int a[n];
  for(int i=0;i<n;i++) cin>>a[i];
  int b[n];
  for(int i=0;i<n;i++) b[i]=a[i];
  sort(b,b+n);
  map<int,int> m;
  for(int i=0;i<n;i++) m[b[i]]=i+1;
  BIT bit(n);
  int ans=0;
  for(int i=0;i<n;i++){
    ans+=i-bit.sum(m[a[i]]);
    bit.add(m[a[i]],1);
  }
  cout<<ans<<endl;
  return 0;
}