1. Abridged Problem Statement
You have n genome strings (lowercase letters, total length up to ~10^6). Then m queries follow; each query gives two strings p and s. For each query, count how many genomes start with p and end with s.

2. Detailed Editorial

Goal
We need to answer queries of the form "how many strings in the dictionary have a given prefix p and suffix s?" quickly, after preprocessing.

Challenges
- Total length of all genomes and queries is large (~10^6).
- Brute-force per query (checking every string) is too slow if done naively.

Key ideas
1. Rolling Hash (Rabin–Karp)
   - Precompute for each dictionary string a prefix-hash array (using a 61-bit Mersenne prime modulus).
   - We also precompute base powers so that we can get hash(s[l..r]) in O(1).

2. Split strings into "small" and "large" by a threshold B (here B=50).
   - Small strings (length < B): there aren't too many possible (prefix, suffix) pairs per string: O(B^2). We enumerate all (prefix, suffix) pairs, compute their hash pairs, and store them in one big array. Sorting that array allows us to count occurrences of any (hash(p), hash(s)) in O(log N).
   - Large strings (length ≥ B): there can be at most total_length/B of them. For each query we simply check each large string in O(1) by comparing its precomputed prefix and suffix hashes to the query hashes.

Complexities
- Preprocessing small strings: ∑L<B(L^2) is manageable since total length is bounded.
- Sorting small-pair list: O(K log K) where K is total small-pairs (~2·10^7).
- Query time: O(log K + #large). #large ≤ total_length/B. For B=50 this is fast enough.

3. C++ Solution

```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;
}
```

4. Python Solution 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)  # H1[i] = hash of s[:i]
    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):
    """
    Get hash of substring [l..r] from prefix array H and power array.
    H has length n+1, power has length n+1.
    """
    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)]
    # Precompute powers up to max total length (~1e6)
    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

    # Separate small and large strings
    small_map = {}  # map ((hp1,hp2),(hs1,hs2)) -> count
    large = []      # list of tuples (len, H1, H2)

    for s in dict_str:
        L = len(s)
        H1, H2 = build_hashes(s)
        if L < B:
            # enumerate all prefix-suffix pairs
            for i in range(L):       # prefix ends at i
                hp1 = get_hash(H1, pow1, 0, i, MOD1)
                hp2 = get_hash(H2, pow2, 0, i, MOD2)
                for j in range(L):   # suffix starts at j
                    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:
            # store for on-the-fly checking
            large.append((L, H1, H2))

    # Process queries
    m = int(input())
    out = []
    for _ in range(m):
        p,suf = input().split()
        lp, ls = len(p), len(suf)
        # build hash of full p and full 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)

        # check each large string in O(1)
        for L,H1_large,H2_large in large:
            if lp > L or ls > L:
                continue
            # prefix check
            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
            # suffix check
            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()
```

5. Compressed Editorial
Use rolling hashes and split strings by length threshold B. For each short string (length < B), enumerate all prefix–suffix hash pairs and store counts in a sorted array. For each query compute its hash pair and look up in the small-string data (binary search), then individually check each long string's prefix and suffix in O(1).
