1. Concise Problem Statement  
Given an array A of N integers (1 ≤ N ≤ 65 537, 0 ≤ Ai ≤ 10^9), count the number of inversion pairs (i, j) such that 1 ≤ i < j ≤ N and A[i] > A[j].

2. Detailed Editorial  

Problem restatement  
You are to count how many pairs of indices (i, j) with i < j satisfy A[i] > A[j]. This is the classic “number of inversions” problem.

Why brute‐force is too slow  
A naive double loop takes O(N²), which for N up to ~65 000 would be ~4 billion operations—far too large for the time limit.

Two standard O(N log N) solutions  
a) Merge‐sort–based counting: During the merge step, whenever you copy an element from the right half before remaining elements in the left half, you add to the inversion count the number of leftover elements in the left.  
b) Fenwick Tree (Binary Indexed Tree, BIT): Sweep from left to right, and for each A[i], count how many of the already‐seen elements are greater than A[i].  

Fenwick Tree approach in detail  
1. Coordinate compression  
   Since Ai can be up to 10^9, but N ≤ 65 537, we compress the values into the range [1..M] where M ≤ N:  
   – Copy all A[i] into a list, sort it, remove duplicates.  
   – Replace each A[i] by its rank in this sorted unique list (1-based).

2. Fenwick Tree structure  
   – Maintain an array fenw[1..M], initially all zeros.  
   – fenw.sum(x) returns the sum of fenw[1]..fenw[x].  
   – fenw.add(x, v) increments fenw[x] by v, updating ancestors.

3. Counting inversions  
   – Initialize inversions = 0.  
   – Iterate i = 0..N−1:  
     • Let r = compressed rank of A[i].  
     • Query small = fenw.sum(r) = number of previous elements ≤ A[i].  
     • Number of previous elements = i.  
     • Then count of previous elements > A[i] = i − small. Add that to inversions.  
     • fenw.add(r, 1) to mark A[i] as seen.

Complexity: O(N log N) time, O(N) extra space.

3. Provided C++ Solution with Detailed Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

// Fast max/min updates
template<class T, class T2>
inline void chkmax(T& x, const T2& y) {
    if (x < y) x = y;
}
template<class T, class T2>
inline void chkmin(T& x, const T2& y) {
    if (x > y) x = y;
}

const int MAXN = (1 << 16);  // up to 65,536

int n;
int a[MAXN + 42];            // input array
vector<int> li;              // for compression

// Fenwick tree variables
int sz;                       // size of compressed range
int tr[MAXN + 42];            // BIT array

// Read input
void read() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
}

// Initialize the Fenwick tree
void init() {
    sz = (int)li.size() + 1;    // we will use indices 1..sz
    memset(tr, 0, sizeof(tr));  // clear BIT
}

// Add +1 at position x in the Fenwick tree
void add(int x) {
    // standard BIT update: climb to parents
    for (; x <= sz; x += (x & -x)) {
        tr[x]++;
    }
}

// Query prefix sum [1..x]
int query(int x) {
    int ret = 0;
    // standard BIT sum: climb to root
    for (; x > 0; x -= (x & -x)) {
        ret += tr[x];
    }
    return ret;
}

// Solve the inversion count
void solve() {
    // 1) Gather all values for compression
    for (int i = 0; i < n; i++) {
        li.push_back(a[i]);
    }
    // 2) Sort and deduplicate
    sort(li.begin(), li.end());
    li.erase(unique(li.begin(), li.end()), li.end());
    // 3) Replace a[i] by its rank in li (1-based)
    for (int i = 0; i < n; i++) {
        a[i] = int(lower_bound(li.begin(), li.end(), a[i]) - li.begin()) + 1;
    }
    // 4) Prepare BIT
    init();
    // 5) Traverse and count inversions
    uint32_t answer = 0;
    for (int i = 0; i < n; i++) {
        // Number of seen elements ≤ a[i]:
        int small = query(a[i]);
        // Total seen is i, so bigger count is i - small
        answer += (i - small);
        // Mark a[i] as seen
        add(a[i]);
    }
    // Output result
    cout << answer << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
import sys
data = sys.stdin.read().strip().split()
n = int(data[0])
A = list(map(int, data[1:]))

# 1. Coordinate compression: map each value to its 1-based rank
vals = sorted(set(A))         # sorted unique values
rank = {v: i+1 for i, v in enumerate(vals)}  # value -> rank
M = len(vals)                 # compressed range size

# 2. Fenwick Tree (BIT) implementation
class Fenwick:
    def __init__(self, size):
        self.n = size
        self.fw = [0] * (size + 1)
    def add(self, i, v):
        # increment position i by v
        while i <= self.n:
            self.fw[i] += v
            i += i & -i
    def sum(self, i):
        # prefix sum [1..i]
        s = 0
        while i > 0:
            s += self.fw[i]
            i -= i & -i
        return s

fenw = Fenwick(M)
inversions = 0

# 3. Traverse the array, count how many previous > current
for i, x in enumerate(A):
    r = rank[x]         # compressed rank
    # number of seen ≤ x is fenw.sum(r)
    smaller_or_equal = fenw.sum(r)
    # seen so far is i, so those > x is i - smaller_or_equal
    inversions += (i - smaller_or_equal)
    # mark this element as seen
    fenw.add(r, 1)

# 4. Print result
print(inversions)
```

5. Compressed Editorial  

- Task: Count pairs i<j with A[i]>A[j] (inversions).  
- N up to 65 537, Ai up to 10^9 ⇒ need O(N log N).  
- Two methods: merge‐sort counting or Fenwick tree.  
- Fenwick approach:  
  1. Coordinate compress A into [1..M].  
  2. Fenwick tree stores counts of seen elements.  
  3. For each A[i], query number ≤ A[i], subtract from i to get count of greater, accumulate.  
  4. Update Fenwick at rank(A[i]).  
- Complexity O(N log N), memory O(N).