<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

505. Prefixes and suffixes
Time limit per test: 0.75 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Andrew has just made a breakthrough in biology: he realized that most of genome's characteristics are determined by the absence or presence of certain prefixes and suffixes. So he needs to quickly test large arrays of genomes for that. More formally, we represent the genome as a string of lowercase English letters. We are given several genomes, and need to answer the following type of question: how many genomes from the given set have the given string p as prefix  the given string s as suffix.
Input
The first line of the input file contains an integer n — the number of genomes. The next n lines contain a non-empty string of lowercase English letters each, representing the given genomes. The total length of all given genomes doesn't exceed . The next line contains an integer m — the number of questions to answer. The next m lines contain two non-empty strings of lowercase English letters each, the first denoting the prefix and the second denoting the suffix. The prefix and suffix are separated with a single space. The total length of all given prefixes plus the total length of all given suffixes doesn't exceed .
Output
Output m integers one per line. i-th output line should contain the number of genomes that have i-th given prefix and i-th given suffix.
Example(s)
sample input
sample output
3
aaaaa
abacabaa
avtobus
6
a a
a aa
aa a
aaaaa aaaa
abac caba
abac a
2
2
1
1
0
1

<|response|>
1. Abridged Problem Statement
Given a collection of n genome strings (total length ≤10^6), answer m queries (total query-string length ≤10^6). Each query gives two strings p (prefix) and s (suffix); output how many genomes start with p and end with s.

2. Key Observations
- We need sub-linear query time after preprocessing.
- Rolling hashes let us compare any prefix or suffix in O(1) after O(L) preprocessing per string.
- If we treat every genome uniformly, per query checking all n genomes is too slow (O(n) per query).
- However, if a genome string is "small" (length < B), then the number of its (prefix, suffix) pairs is O(B^2). Summed over all small strings this can be affordable if B is chosen moderately (e.g. B=50). We can enumerate all such pairs once and store them in a big array or hash-map with counts.
- "Large" genomes (length ≥ B) are few—at most (total_length)/B—and we can check each of them in O(1) per query using hash comparisons.

3. Full Solution Approach
1. Choose a threshold B (e.g. B = 50).
2. Precompute a rolling-hash base and powers up to the maximum total string length (~10^6).
3. For each genome string sᵢ:
   a. Build its prefix-hash array Hᵢ.
   b. If |sᵢ| < B (small string), enumerate all pairs (prefix_end, suffix_start) and compute the hash of sᵢ[0..prefix_end] and sᵢ[suffix_start..|sᵢ|−1], then store the pair of hashes in a vector `small_pairs`.
   c. Otherwise (large string), record its index in `large_indices` for on-the-fly checking.
4. Sort the vector `small_pairs`. This lets us, for any given (hp, hs), count how many small-string pairs match in O(log N) via binary search.
5. For each query (p, s):
   a. Compute hash hp of p and hash hs of s.
   b. Count matches among small strings by finding the range of equal (hp, hs) in `small_pairs`.
   c. For each large string index j in `large_indices`, verify in O(1) that its first |p| characters hash to hp and its last |s| characters hash to hs. Add these matches.
6. Output the total count.

Time & Memory Complexity
- Preprocessing small strings: ∑_{|sᵢ|<B} O(|sᵢ|^2) ≤ O(total_length·B).
- Sorting small_pairs: O(K log K), K≈∑|sᵢ|^2 for small strings.
- Query time: O(log K + (#large_strings)). #large_strings ≤ total_length/B.
Choosing B≈50 balances these terms under the given constraints.

4. C++ Implementation with Detailed Comments
```cpp
#include <bits/stdc++.h>
// #include <coding_library/strings/hashing.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;
};

class HashMeta {
  private:
    void set_random_base() {
        seed_seq seed{
            (uint32_t)chrono::duration_cast<chrono::nanoseconds>(
                chrono::high_resolution_clock::now().time_since_epoch()
            )
                .count(),
            (uint32_t)random_device()(), (uint32_t)42
        };
        mt19937 rng(seed);
        base = uniform_int_distribution<uint64_t>(0, mod - 1)(rng);
    }

    void precompute_base_pow(size_t n) {
        base_pow.resize(n);
        base_pow[0] = 1;
        for(size_t i = 1; i < n; i++) {
            base_pow[i] = mul(base_pow[i - 1], base);
        }
    }

    static constexpr uint64_t add(uint64_t a, uint64_t b) {
        a += b + 1;
        a = (a & mod) + (a >> 61);
        return a - 1;
    }

    static constexpr uint64_t sub(uint64_t a, uint64_t b) {
        return add(a, mod - b);
    }

    static constexpr uint64_t mul(uint64_t a, uint64_t b) {
        uint64_t l1 = (uint32_t)a, h1 = a >> 32, l2 = (uint32_t)b, h2 = b >> 32;
        uint64_t l = l1 * l2, m = l1 * h2 + l2 * h1, h = h1 * h2;
        uint64_t ret =
            (l & mod) + (l >> 61) + (h << 3) + (m >> 29) + (m << 35 >> 3) + 1;
        ret = (ret & mod) + (ret >> 61);
        ret = (ret & mod) + (ret >> 61);
        return ret - 1;
    }

  public:
    class hash_t {
        uint64_t h;

      public:
        hash_t() : h(0) {}
        hash_t(uint64_t h) : h(h) {}
        operator uint64_t() const { return h; }

        hash_t& operator+=(const hash_t& other) {
            h = add(h, other.h);
            return *this;
        }

        hash_t& operator-=(const hash_t& other) {
            h = sub(h, other.h);
            return *this;
        }

        hash_t& operator*=(const hash_t& other) {
            h = mul(h, other.h);
            return *this;
        }

        hash_t operator+(const hash_t& other) const {
            return hash_t(*this) += other;
        }
        hash_t operator-(const hash_t& other) const {
            return hash_t(*this) -= other;
        }
        hash_t operator*(const hash_t& other) const {
            return hash_t(*this) *= other;
        }

        bool operator==(const hash_t& other) const { return h == other.h; }
        bool operator!=(const hash_t& other) const { return h != other.h; }

        // For use in std::map and std::set
        bool operator<(const hash_t& other) const { return h < other.h; }
    };

    uint64_t base;
    vector<hash_t> base_pow;
    static constexpr uint64_t mod = (1ull << 61) - 1;

    void init(size_t n) {
        set_random_base();
        precompute_base_pow(n);
    }

    vector<hash_t> rabin_karp(const string& s) {
        vector<hash_t> h(s.size());
        for(size_t i = 0; i < s.size(); i++) {
            h[i] = (i ? h[i - 1] : hash_t(0)) * hash_t(base) + hash_t(s[i]);
        }
        return h;
    }

    hash_t hash_range(int l, int r, const vector<hash_t>& h) {
        if(l == 0) {
            return h[r];
        }
        return h[r] - h[l - 1] * base_pow[r - l + 1];
    }
};

HashMeta hash_meta;
using hash_t = HashMeta::hash_t;
const int B = 50;

int n;
vector<string> dictionary;

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

void solve() {
    // Count dictionary words that start with a given prefix and end with a
    // given suffix, using a square-root split on word length.
    //
    // Threshold B: a word is "small" if its length is below B, "large"
    // otherwise. There are at most total_len / B large words, so few of
    // them.
    //
    // For each small word we enumerate every (prefix, suffix) pair of its
    // own and store the pair of polynomial hashes; sorting this list lets a
    // query answer the small-word contribution by counting the matching
    // (prefix_hash, suffix_hash) pair via binary search.
    //
    // For each query we additionally test the few large words directly: a
    // large word matches if its length is at least max(|p|, |s|) and the
    // hash of its leading |p| chars equals the prefix hash and the hash of
    // its trailing |s| chars equals the suffix hash.

    vector<pair<hash_t, hash_t>> small_hashes;

    vector<int> large_strings;
    vector<vector<hash_t>> hashes(n);
    for(int k = 0; k < n; k++) {
        hashes[k] = hash_meta.rabin_karp(dictionary[k]);
        if(dictionary[k].size() >= B) {
            large_strings.push_back(k);
        } else {
            for(int i = 0; i < (int)hashes[k].size(); i++) {
                for(int j = 0; j < (int)hashes[k].size(); j++) {
                    hash_t phash = hash_meta.hash_range(0, i, hashes[k]);
                    hash_t shash = hash_meta.hash_range(
                        j, hashes[k].size() - 1, hashes[k]
                    );
                    small_hashes.push_back({phash, shash});
                }
            }
        }
    }

    sort(small_hashes.begin(), small_hashes.end());

    int q;
    cin >> q;
    while(q--) {
        string req_pref, req_suff;
        cin >> req_pref >> req_suff;
        auto hp = hash_meta.rabin_karp(req_pref);
        auto hs = hash_meta.rabin_karp(req_suff);

        int ans = upper_bound(
                      small_hashes.begin(), small_hashes.end(),
                      make_pair(hp.back(), hs.back())
                  ) -
                  lower_bound(
                      small_hashes.begin(), small_hashes.end(),
                      make_pair(hp.back(), hs.back())
                  );
        for(int k: large_strings) {
            if(dictionary[k].size() < max(req_pref.size(), req_suff.size())) {
                continue;
            }
            hash_t phash =
                hash_meta.hash_range(0, (int)req_pref.size() - 1, hashes[k]);
            hash_t shash = hash_meta.hash_range(
                (int)dictionary[k].size() - req_suff.size(),
                (int)dictionary[k].size() - 1, hashes[k]
            );
            if(phash == hp.back() && shash == hs.back()) {
                ans++;
            }
        }

        cout << ans << '\n';
    }
}

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

    hash_meta.init((int)1e6);

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

    return 0;
}
```

5. Python Implementation with Detailed Comments
```python
import sys
sys.setrecursionlimit(10**7)
input = sys.stdin.readline

# We'll use double hashing to reduce collision risk
MOD1, MOD2 = 10**9+7, 10**9+9
BASE1, BASE2 = 91138233, 97266353
B = 50   # threshold between small and large

def build_hashes(s):
    """Return two prefix-hash arrays H1, H2 for string s."""
    n = len(s)
    H1 = [0]*(n+1)
    H2 = [0]*(n+1)
    for i,ch in enumerate(s):
        v = ord(ch) - ord('a') + 1
        H1[i+1] = (H1[i]*BASE1 + v) % MOD1
        H2[i+1] = (H2[i]*BASE2 + v) % MOD2
    return H1, H2

def get_hash(H, power, l, r, mod):
    return (H[r+1] - H[l]*power[r-l+1] % mod + mod) % mod

def main():
    n = int(input())
    dict_str = [input().strip() for _ in range(n)]
    maxlen = 10**6 + 5
    pow1 = [1]* (maxlen)
    pow2 = [1]* (maxlen)
    for i in range(1, maxlen):
        pow1[i] = pow1[i-1]*BASE1 % MOD1
        pow2[i] = pow2[i-1]*BASE2 % MOD2

    small_map = {}
    large = []

    for s in dict_str:
        L = len(s)
        H1, H2 = build_hashes(s)
        if L < B:
            for i in range(L):
                hp1 = get_hash(H1, pow1, 0, i, MOD1)
                hp2 = get_hash(H2, pow2, 0, i, MOD2)
                for j in range(L):
                    hs1 = get_hash(H1, pow1, j, L-1, MOD1)
                    hs2 = get_hash(H2, pow2, j, L-1, MOD2)
                    key = ((hp1,hp2),(hs1,hs2))
                    small_map[key] = small_map.get(key, 0) + 1
        else:
            large.append((L, H1, H2))

    m = int(input())
    out = []
    for _ in range(m):
        p,suf = input().split()
        lp, ls = len(p), len(suf)
        Hp1, Hp2 = build_hashes(p)
        Hs1, Hs2 = build_hashes(suf)
        key = ((Hp1[-1],Hp2[-1]), (Hs1[-1],Hs2[-1]))
        ans = small_map.get(key, 0)

        for L,H1_large,H2_large in large:
            if lp > L or ls > L:
                continue
            ph1 = get_hash(H1_large, pow1, 0, lp-1, MOD1)
            ph2 = get_hash(H2_large, pow2, 0, lp-1, MOD2)
            if (ph1,ph2) != (Hp1[-1],Hp2[-1]):
                continue
            sh1 = get_hash(H1_large, pow1, L-ls, L-1, MOD1)
            sh2 = get_hash(H2_large, pow2, L-ls, L-1, MOD2)
            if (sh1,sh2) == (Hs1[-1],Hs2[-1]):
                ans += 1
        out.append(str(ans))

    print('\n'.join(out))

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

Explanation of Steps
- We use rolling hashes so that any prefix or suffix hash can be obtained in O(1).
- We split strings by length B to trade preprocessing for query speed.
- All short-string (prefix, suffix) pairs are pre-enumerated and counted.
- Large strings are few, so checking them per query is still efficient. This achieves an overall fast solution under the input size constraints.
