1. Abridged Problem Statement  
Given two integers a and b (2 ≤ a, b ≤ 10^6), define the key number of n as follows: let its distinct prime divisors be p₁, p₂, …, p_k; let M = max(p_i) and S = ∑ p_i. Then key(n) = M − (S − M) = 2·M − S. The input guarantees key(a) ≠ key(b). Print “a” if key(a) > key(b); otherwise print “b.”

2. Detailed Editorial  
Problem restated  
We must compare two specially defined values (key numbers) for a and b, and output which one is larger.

Key number definition  
- Extract all distinct prime factors of n: {p₁, p₂, …, p_k}.  
- Let M = largest prime in that set.  
- Let S = sum of all primes in the set.  
- Define key(n) = M − (S − M) = 2·M − S.

Constraints and complexity  
- a and b up to 10^6.  
- We only have two numbers to factor, so a simple trial‐division up to √n (≤1000) is perfectly fine.  
- Worst‐case cost: O(√n) per number, i.e. ~2·10^3 operations, negligible.

Prime factorization approach  
1. Initialize an empty list of primes.  
2. For i from 2 to ⌊√n⌋:  
   - If i divides n, record i in the list, then divide n by i repeatedly until it no longer divides.  
3. If after that n > 1, the remainder is itself prime; record it.  
4. You get a list of distinct prime factors.

Computing the key number in one pass  
- While collecting primes, track their sum S and current maximum M.  
- Finally compute key = 2·M − S.

Putting it all together  
- Read a, b.  
- Compute key(a), key(b).  
- Compare and print “a” or “b.”

3. Provided C++ Solution with Detailed Comments  
#include <bits/stdc++.h>  
using namespace std;

// Overload << for pair so we can print pairs easily (not used here, but provided)  
template<typename T1, typename T2>  
ostream &operator<<(ostream &out, const pair<T1, T2> &x) {  
    return out << x.first << ' ' << x.second;  
}

// Overload >> for pair so we can read pairs easily (not used here)  
template<typename T1, typename T2>  
istream &operator>>(istream &in, pair<T1, T2> &x) {  
    return in >> x.first >> x.second;  
}

// Overload >> for vector so we can read a list of values in one statement (not used here)  
template<typename T>  
istream &operator>>(istream &in, vector<T> &a) {  
    for(auto &x: a) {  
        in >> x;  
    }  
    return in;  
}

// Overload << for vector so we can print a list of values easily (not used here)  
template<typename T>  
ostream &operator<<(ostream &out, const vector<T> &a) {  
    for(auto x: a) {  
        out << x << ' ';  
    }  
    return out;  
}

// Returns the distinct prime factors of n in a vector  
vector<int> prime_factors(int n) {  
    vector<int> res;  
    // Try dividing by every i from 2 up to sqrt(n)  
    for(int i = 2; i * i <= n; i++) {  
        if(n % i == 0) {  
            // i is a prime factor; record it once  
            res.push_back(i);  
            // Remove all powers of i from n  
            while(n % i == 0) {  
                n /= i;  
            }  
        }  
    }  
    // If n > 1, what remains is a prime > sqrt(original n)  
    if(n > 1) {  
        res.push_back(n);  
    }  
    return res;  
}

// Compute the key number: 2*max(primes) - sum(primes)  
int f(int x) {  
    vector<int> p = prime_factors(x);  
    // Sum up all distinct primes  
    int64_t sum = accumulate(p.begin(), p.end(), 0LL);  
    // Find the largest prime  
    int mx = *max_element(p.begin(), p.end());  
    // Return 2*mx - sum  
    return 2 * mx - sum;  
}

int a, b;

// Read input values a and b  
void read() {  
    cin >> a >> b;  
}

// Decide which door to open by comparing key(a) and key(b)  
void solve() {  
    // If key(a) > key(b), the answer is "a", else "b"  
    if(f(a) > f(b)) {  
        cout << "a\n";  
    } else {  
        cout << "b\n";  
    }  
}

int main() {  
    ios_base::sync_with_stdio(false);  // Speeds up I/O  
    cin.tie(nullptr);                  // Untie cin/cout  

    int T = 1;  
    // If there were multiple test cases, we'd read T here  
    for(int test = 1; test <= T; test++) {  
        read();  
        solve();  
    }  

    return 0;  
}

4. Python Solution with Detailed Comments  
```python
import sys
import math

def prime_factors(n):
    """Return the distinct prime factors of n."""
    factors = []
    # Check divisibility by 2 separately for speed (optional)
    if n % 2 == 0:
        factors.append(2)
        while n % 2 == 0:
            n //= 2
    # Now n is odd; test odd divisors from 3 up to sqrt(n)
    i = 3
    # Stop when i*i > n
    while i * i <= n:
        if n % i == 0:
            factors.append(i)
            # Remove all copies of i
            while n % i == 0:
                n //= i
        i += 2
    # If remainder > 1, it's a prime
    if n > 1:
        factors.append(n)
    return factors

def key_number(n):
    """Compute key(n) = 2*max(primes) - sum(primes)."""
    primes = prime_factors(n)
    s = sum(primes)
    m = max(primes)
    return 2*m - s

def main():
    # Read two integers from stdin
    a, b = map(int, sys.stdin.readline().split())
    # Compare keys and print result
    if key_number(a) > key_number(b):
        print("a")
    else:
        print("b")

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

5. Compressed Editorial  
- Factor each of a and b into distinct primes via trial division up to √n.  
- Let S = sum(distinct primes), M = max(distinct primes). Key = 2·M − S.  
- Compute key(a) and key(b); output “a” if key(a) > key(b), else “b.”