1. Abridged 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 tr[1..M], initially all zeros.  
   – query(x) returns the sum of tr[1]..tr[x].  
   – add(x) increments tr[x] by 1, updating ancestors.

3. Counting inversions  
   – Initialize answer = 0.  
   – Iterate i = 0..N−1:  
     • Let r = compressed rank of A[i].  
     • Insert it first with add(r); now i+1 elements have been seen.  
     • query(r) is the number of seen elements ≤ A[i] (including itself).  
     • The remaining i+1 − query(r) seen elements are strictly greater and sit to the left, i.e. inversions ending at position i. Add that to answer.

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

3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

class Fenwick {
  public:
    int n;
    vector<int> tr;

    Fenwick(int _n = 0) { init(_n); }

    void init(int _n) {
        n = _n;
        tr.assign(n + 1, 0);
    }

    void add(int x) {
        for(; x <= n; x += (x & -x)) {
            tr[x]++;
        }
    }

    int query(int x) {
        int ret = 0;
        for(; x > 0; x -= (x & -x)) {
            ret += tr[x];
        }

        return ret;
    }
};

int n;
vector<int> a;

void read() {
    cin >> n;
    a.resize(n);
    cin >> a;
}

void solve() {
    // Count inversions with a Fenwick tree over compressed values. We sort and
    // dedupe the values, map each element to its 1-based rank, then sweep left
    // to right inserting ranks. After inserting the i-th element (i + 1 seen so
    // far), the number of already-inserted values that are <= a[i] is
    // query(a[i]); the rest, i + 1 - query(a[i]), are strictly greater values
    // sitting to its left, i.e. inversions ending at position i.

    vector<int> li = a;
    sort(li.begin(), li.end());
    li.erase(unique(li.begin(), li.end()), li.end());

    for(int i = 0; i < n; i++) {
        a[i] = lower_bound(li.begin(), li.end(), a[i]) - li.begin() + 1;
    }

    Fenwick fen(li.size() + 1);

    uint32_t answer = 0;
    for(int i = 0; i < n; i++) {
        fen.add(a[i]);
        answer += i + 1 - fen.query(a[i]);
    }

    cout << answer << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution

```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], the seen elements greater than A[i] are inversions; accumulate them.  
  4. Update Fenwick at rank(A[i]).  
- Complexity O(N log N), memory O(N).
