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;

typedef long long ll;
typedef vector<int> vi;
typedef vector<string> vs;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define rep(i, n)    FOR(i, 0, n)
#define pb           push_back
#define all(a)       (a).begin(), (a).end()
#define UNIQUE(v)    v.erase(unique(all(v)), v.end());
#define fill(a, x)   memset(a, x, sizeof(a))
#define zero(a)      fill(a, 0)
#define minus(a)     fill(a, -1)
#define dbg(x)       cout << #x "=" << x << endl
#define MAX          200000
#define SENTINEL     2000000000

int L[MAX / 2 + 2], R[MAX / 2 + 2];

ll merge(int A[], int n, int left, int mid, int right) {
  int i, j, k;
  ll cntC = 0;
  int n1 = mid - left;
  int n2 = right - mid;

  for (i = 0; i < n1; i++) {
    L[i] = A[left + i];
  }

  for (i = 0; i < n2; i++) {
    R[i] = A[mid + i];
  }

  L[n1] = R[n2] = SENTINEL;
  i = j = 0;

  for (k = left; k < right; k++) {
    if (L[i] <= R[j]) {
      A[k] = L[i++];
    } else {
      A[k] = R[j++];
      cntC += n1 - i;
    }
  }
  return cntC;
}

ll mergeSort(int A[], int n, int left, int right) {
  int mid;
  ll v1, v2, v3;

  if (left + 1 < right) {
    mid = (left + right) / 2;
    v1 = mergeSort(A, n, left, mid);
    v2 = mergeSort(A, n, mid, right);
    v3 = merge(A, n, left, mid, right);
    return v1 + v2 + v3;
  } else {
    return 0;
  }
}

int main() {
  int A[MAX], n, i;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> A[i];
  }
  ll ans = mergeSort(A, n, 0, n);
  cout << ans << endl;
  return 0;
}