1. Abridged Problem Statement  
Given an integer N (1 ≤ N ≤ 10^6), find all pairs of primes (A, B) with A ≤ B such that A + B is also prime and does not exceed N. First output the number of such pairs, then list each pair on its own line.

2. Detailed Editorial  

Overview  
We must list all prime pairs (A, B) so that A + B is prime and ≤ N. A direct double loop over primes would be O(π(N)^2) and too slow for N up to 10^6. Instead, observe:

Key Observation  
– Aside from 2, every prime is odd.  
– The sum of two odd primes is even and, if greater than 2, not prime.  
– Therefore, to have A + B prime, one of A or B must be 2 (the only even prime).  

So all candidate pairs are of the form (2, p) with p prime, and 2 + p must itself be prime and ≤ N. That means p + 2 must be prime and ≤ N.

Algorithm Steps  
1. Sieve of Eratosthenes up to N: build a bitset `isComposite[]` where `isComposite[x]` is true if x is not prime.  
2. Iterate p from 2 to N–2:  
   – Check if p is prime (`!isComposite[p]`)  
   – Check if p + 2 is prime (`!isComposite[p+2]`)  
   – If both true, record the pair (2, p).  
3. Output the total count and the list of pairs.

Complexity  
– Sieve: O(N log log N) with a classic implementation, or O(N log N) in this marking style.  
– Checking p and p + 2 for each p up to N is O(N).  
Total fits comfortably for N ≤ 10^6.

3. Provided C++ Solution with Line-by-Line Comments  

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

// A bitset large enough for N up to 1e6
const int MAXN = (1 << 20);

int n;                     // Input upper bound
bitset<MAXN> isComposite;  // Marks composite numbers: true if not prime

// Read n and build a simple sieve
void read() {
    cin >> n;
    // 0 and 1 are not prime
    isComposite[0] = isComposite[1] = 1;
    // Mark multiples: for each x from 2..n, mark 2x,3x,... as composite
    for (int x = 2; x <= n; x++) {
        for (int y = 2 * x; y <= n; y += x) {
            isComposite[y] = 1;
        }
    }
}

// Find and output all prime pairs (2, p) such that p and p+2 are primes
void solve() {
    vector<pair<int,int>> ans;
    // We only need to check p up to n-2 so that p+2 ≤ n
    for (int p = 2; p + 2 <= n; p++) {
        // if p is prime and p+2 is prime, we have a valid pair (2,p)
        if (!isComposite[p] && !isComposite[p + 2]) {
            ans.emplace_back(2, p);
        }
    }
    // Output the number of pairs
    cout << ans.size() << '\n';
    // Output each pair on its own line
    for (auto &pr : ans) {
        cout << pr.first << ' ' << pr.second << '\n';
    }
}

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

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

4. Python Solution with Detailed Comments  

```python
import sys

def main():
    data = sys.stdin.read().strip().split()
    n = int(data[0])

    # Sieve array: False means prime, True means composite
    is_composite = [False] * (n + 1)
    if n >= 0:
        is_composite[0] = True
    if n >= 1:
        is_composite[1] = True

    # Build sieve: mark composites
    for x in range(2, n + 1):
        if not is_composite[x]:
            # Mark multiples of x as composite
            for y in range(2 * x, n + 1, x):
                is_composite[y] = True

    result = []
    # Only pairs of the form (2, p) can sum to another prime
    # Check p from 2 to n-2
    for p in range(2, n - 1):
        if not is_composite[p] and not is_composite[p + 2]:
            # p is prime and p+2 is prime => 2+p is prime
            result.append((2, p))

    # Output
    out = [str(len(result))]
    for a, b in result:
        out.append(f"{a} {b}")
    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
We need prime pairs (A,B) with A+B prime. Because any two odd primes sum to an even >2 (not prime), one prime must be 2. Thus look for primes p where p and p+2 are both prime (twin primes). Sieve up to N in O(N log log N) and scan p from 2 to N–2, collecting (2,p) whenever both p and p+2 are prime. Output the count and the list.