1. Abridged Problem Statement
Given a binary string S of length N (characters ‘a’ and ‘b’), find the shortest nonempty binary string over {‘a’,‘b’} that does *not* occur in S as a contiguous substring. Output its length and one such string.

2. Detailed Editorial
Goal: find the shortest absent substring over alphabet {a,b}.
Observation: If you look at all substrings of S of length L, there are at most N−L+1 such substrings but 2^L possible binary strings of length L. As soon as 2^L > N−L+1, by pigeonhole principle there is some string of length L missing. In practice the answer L will be quite small (≤ about log₂N+1, e.g. ≤20 when N=5⋅10^5).

Algorithm outline:
1. For L = 1, 2, 3, …
2.   Use a sliding‐window bitmask of width L to record which of the 2^L masks appear in S.
     - Map ‘a’→0, ‘b’→1.
     - Initialize mask to the integer value of S[0..L−1].
     - Mark `cnt[mask]++`.
     - For i from L to N−1:
         • Shift mask left by 1 bit, mask &= (2^L−1) to drop the high bit, then add bit for S[i].
         • Mark `cnt[mask]++`.
3.   After scanning, scan all masks from 0 to 2^L−1; if any mask has count 0, that mask encodes an absent string of length L.
4.   Reconstruct the string from the mask by reading its bits from most significant to least, mapping 0→‘a’, 1→‘b’.
5.   Output L and the reconstructed string; terminate.

Complexity: for each L we do O(N) work plus O(2^L) to check which masks appear. Since L grows until 2^L > N, Lmax ≈ log₂N+1 (≈20), the total is O(N·Lmax + 2^Lmax) = O(N log N + N), acceptable for N≤5⋅10^5.

3. C++ Solution
```cpp
#include <bits/stdc++.h>
// #include <coding_library/strings/suffix_automaton.hpp>

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;
};

int n;
string s;

void read() { cin >> n >> s; }

void solve() {
    // The shortest absent substring has length at most ~20 since a string of
    // length n contains at most n distinct substrings of any given length, so
    // once 2^len > n some length-len word must be missing. We try increasing
    // lengths len = 1, 2, ...; for each we slide a window over the key-string
    // encoding the current window as a binary mask (a -> 0, b -> 1) and tally
    // cnt[mask] for every length-len occurrence. The first mask with count
    // zero is an absent word; we print it most-significant bit first.

    vector<int> cnt;
    for(int len = 1;; len++) {
        cnt.assign(1 << len, 0);
        int mask = 0;
        for(int i = 0; i < len; i++) {
            mask = (mask << 1) | (s[i] - 'a');
        }
        cnt[mask]++;
        for(int i = len; i < n; i++) {
            mask = ((mask << 1) & ((1 << len) - 1)) | (s[i] - 'a');
            cnt[mask]++;
        }

        for(int i = 0; i < (1 << len); i++) {
            if(cnt[i] == 0) {
                cout << len << '\n';
                for(int j = len - 1; j >= 0; j--) {
                    cout << (char)((i >> j & 1) + 'a');
                }
                return;
            }
        }
    }
}

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 with Detailed Comments
```python
def find_shortest_absent(s):
    n = len(s)
    # Map characters to bits 0 or 1
    bits = [ord(c) - ord('a') for c in s]

    # Try each length L = 1, 2, 3, ...
    L = 1
    while True:
        # There are 2^L possible patterns; track which ones we see
        size = 1 << L
        seen = [False] * size

        # Build initial mask from first L bits
        mask = 0
        for i in range(L):
            mask = (mask << 1) | bits[i]
        seen[mask] = True

        # Slide window over the rest; maintain rolling hash in 'mask'
        full_mask = size - 1
        for i in range(L, n):
            mask = ((mask << 1) & full_mask) | bits[i]
            seen[mask] = True

        # Find first mask not seen
        for m in range(size):
            if not seen[m]:
                # reconstruct the string of length L
                res = []
                for b in range(L - 1, -1, -1):
                    bit = (m >> b) & 1
                    res.append(chr(bit + ord('a')))
                return L, ''.join(res)

        # not found at this length, try next
        L += 1

if __name__ == "__main__":
    import sys
    data = sys.stdin.read().split()
    n = int(data[0])
    s = data[1].strip()
    length, keyword = find_shortest_absent(s)
    print(length)
    print(keyword)
```

Explanation of Python code steps:
- Convert input string to a list of 0/1 bits.
- For each candidate length L, create a boolean array `seen` of size 2^L.
- Compute the initial window mask, then iterate with a rolling update.
- Mark every seen mask.
- Scan for the first unseen mask, reconstruct the string, and return it.

5. Compressed Editorial
We must find the shortest string over {a,b} missing from S. Since there are 2^L possible length-L strings but only N−L+1 substrings of that length in S, at some L≲log₂N+1 a missing one exists. For each L starting at 1, use a bitmask sliding window of width L to mark all substrings seen; then scan masks 0…2^L−1 for the first unvisited. Reconstruct that mask to the answer string and output L and the string. This runs in O(N·L_max + 2^L_max) ≈ O(N log N).
