## 1) Abridged problem statement

You are given a nickname consisting of Latin letters, digits, and spaces, plus integers `q` and `c` (`1 ≤ q ≤ c ≤ 63`). After a rule change, users may also use Russian letters that look identical to some Latin letters.

Ambiguous (look-alike) Latin letters are:
`A B C E H K M O P T X a c e o p x y` (18 letters).
Each occurrence of such a character can be replaced by its Russian analogue (or vice versa), independently.

Nicknames are split into **words** (maximal substrings separated by spaces). You may:
- Replace any subset of ambiguous letters by their analogues.
- Change the number of spaces **between words**, and add/remove leading/trailing spaces,
  but:
  - you cannot remove all spaces between two neighboring words (words must stay separated),
  - you cannot insert spaces inside a word.

Count how many different "mult" nicknames exist whose total length is between `q` and `c` inclusive. Do not count the original nickname itself.

Output the count (it can be very large).

---

## 2) Detailed editorial (solution explanation)

### Observations

The transformations affect the nickname in two independent ways:

1) **Letter substitutions**
Only the 18 "ambiguous" Latin letters can be swapped with a Russian look-alike.
If the nickname contains `A` ambiguous characters total (across all words), then each can be chosen to flip or not flip, independently:
- number of letter variants = `2^A`.

Digits and non-ambiguous letters have only 1 option each.

2) **Spaces redistribution**
Let the nickname contain `w` words (split by one or more spaces). Words themselves cannot change or merge/split; only spaces between words, plus leading/trailing spaces, can change.

Let:
- `L` = total number of non-space characters (sum of word lengths).
- `S` = total number of spaces in the resulting nickname.
Then the final length is `L + S`, which must satisfy:
- `q ≤ L + S ≤ c`.

Also, the constraints on spaces:
- Between every adjacent pair of words, there must be **at least 1** space (cannot delete all separators).
- Leading and trailing spaces can be **0 or more**.
- No spaces can appear inside words (already enforced by keeping words intact).

So for a fixed `S`, we need to count the number of ways to distribute `S` spaces into:
- `(w-1)` mandatory "gaps" between words, each ≥ 1,
- 1 leading gap ≥ 0,
- 1 trailing gap ≥ 0.

That is `w+1` gaps total:
`lead, gap1, gap2, ..., gap(w-1), trail`
with constraints:
- `gap_i ≥ 1` for `i=1..w-1`
- `lead, trail ≥ 0`
- sum of all gaps = `S`.

Convert to nonnegative variables by subtracting 1 from each internal gap:
Let `gap'_i = gap_i - 1 ≥ 0`. Then
`lead + trail + sum gap'_i = S - (w-1)`.

Number of nonnegative solutions to:
`x1 + x2 + ... + x_{w+1} = S - (w-1)`
(where `w-1` internal gaps are `gap'`, and 2 are lead/trail)
is a classic stars-and-bars count:
\[
\binom{(S-(w-1)) + (w+1) - 1}{(w+1) - 1}
= \binom{S+1}{w}
\]
So for a fixed total spaces `S`, the number of spacing variants is `C(S+1, w)`.

### Valid range of spaces

We need `q ≤ L + S ≤ c` ⇒ `q - L ≤ S ≤ c - L`.

Additionally, since between words we need at least `w-1` spaces:
`S ≥ w-1`.

And of course `S ≥ 0`.

So:
- `S_min = max(0, w-1, q-L)`
- `S_max = c-L`

If `S_min > S_max`, then there are no valid nicknames (answer 0).

### Total count

For each `S` in `[S_min, S_max]`:
- spacing ways = `C(S+1, w)`

Total spacing ways:
\[
\text{space\_ways} = \sum_{S=S_{\min}}^{S_{\max}} \binom{S+1}{w}
\]

Total nicknames including the original:
\[
\text{total} = 2^A \cdot \text{space\_ways}
\]

We must **exclude the original nickname itself**, which is always counted once (choose no letter flips and keep original spacing). Therefore:
\[
\text{answer} = 2^A \cdot \text{space\_ways} - 1
\]

### Big integers

The result can be huge (length up to 63, but combinatorics and `2^A` can be large), so we need arbitrary-precision arithmetic (big integers).

The provided C++ code includes a `bigint` implementation (base 1e9, Karatsuba multiplication).

### Computing combinations efficiently

We need `C(S+1, w)` for small `w` (number of words) and `S ≤ 63`.
The code computes it via multiplicative formula:
\[
\binom{n}{w} = \prod_{j=0}^{w-1} \frac{n-j}{j+1}
\]
with `n = S+1`, performing exact division at each step using big integers.

Complexities are tiny: `S` iterates at most 64 values; `w` at most 64.

---

## 3) C++ Solution

```cpp
#include <bits/stdc++.h>

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;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

int q, c;
string s;

void read() {
    cin >> q >> c;
    getline(cin, s);
    getline(cin, s);
}

void solve() {
    // A mult is obtained by independently (a) swapping any subset of the
    // ambiguous Latin letters for their Russian look-alike, and (b) choosing
    // how many spaces separate consecutive words and how many leading/trailing
    // spaces to add, keeping total length in [q, c] and at least one space
    // between adjacent words.
    //
    // - letter_ways = 2^(number of ambiguous letters): each such letter is
    //   independently kept Latin or made Russian.
    //
    // - For a fixed number of space characters sp, the words plus their gaps are
    //   placed into sp + 1 "slots" (before/between/after words) with each of the
    //   w - 1 internal gaps holding at least one space. The count of such
    //   distributions is C(sp + 1, w), summed over sp from max(0, w - 1,
    //   q - total_len) to c - total_len.
    //
    // The total number of nicknames (including the original) is letter_ways *
    // space_ways, computed with the vendored bigint; we subtract one to exclude
    // the original nickname itself.

    const string ambiguous = "ABCEHKMOPTXaceopxy";
    set<char> amb_set(ambiguous.begin(), ambiguous.end());

    vector<string> words;
    for(int i = 0; i < (int)s.size();) {
        if(s[i] == ' ') {
            i++;
            continue;
        }
        int j = i;
        while(j < (int)s.size() && s[j] != ' ') {
            j++;
        }
        words.push_back(s.substr(i, j - i));
        i = j;
    }

    int w = (int)words.size();
    int total_len = 0;
    int amb_cnt = 0;
    for(auto& word: words) {
        total_len += (int)word.size();
        for(char ch: word) {
            if(amb_set.count(ch)) {
                amb_cnt++;
            }
        }
    }

    bigint letter_ways(1);
    for(int i = 0; i < amb_cnt; i++) {
        letter_ways *= 2;
    }

    int s_min = max({0, w - 1, q - total_len});
    int s_max = c - total_len;

    bigint space_ways(0);
    for(int sp = s_min; sp <= s_max; sp++) {
        bigint comb(1);
        for(int j = 0; j < w; j++) {
            comb *= (sp + 1 - j);
            comb /= (j + 1);
        }
        space_ways += comb;
    }

    cout << letter_ways * space_ways - bigint(1) << "\n";
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (with detailed comments)

```python
import sys
import math

def main() -> None:
    # Read q, c
    line1 = sys.stdin.readline().strip()
    q, c = map(int, line1.split())

    # Read nickname line (may contain spaces, may be empty but per statement it's a nickname)
    s = sys.stdin.readline().rstrip("\n")

    # Set of Latin letters that have Russian look-alikes
    amb = set("ABCEHKMOPTXaceopxy")

    # Split into words by spaces (consecutive spaces collapse)
    # This matches the C++ manual split behavior.
    words = [w for w in s.split(' ') if w != '']
    w = len(words)

    # Total non-space length L and ambiguous count A
    L = sum(len(word) for word in words)
    A = sum(1 for word in words for ch in word if ch in amb)

    # Letter variants: 2^A
    letter_ways = 1 << A  # Python big int

    # Valid total spaces S range
    s_min = max(0, w - 1, q - L)
    s_max = c - L

    if s_min > s_max:
        print(0)
        return

    # For fixed S, spacing variants = C(S+1, w)
    # Sum these over S in [s_min, s_max].
    space_ways = 0
    for S in range(s_min, s_max + 1):
        space_ways += math.comb(S + 1, w)

    # Total = letter_ways * space_ways, exclude the original nickname itself
    ans = letter_ways * space_ways - 1
    print(ans)

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

---

## 5) Compressed editorial

- Split nickname into `w` words (by spaces). Let `L` be total non-space length.
- Count `A`: occurrences of ambiguous letters in the given list. Letter variants = `2^A`.
- Let `S` be total spaces in the final nickname. Constraints:
  - `q ≤ L+S ≤ c` ⇒ `q-L ≤ S ≤ c-L`
  - separators between words require `S ≥ w-1`
  - also `S ≥ 0`
  So `S ∈ [max(0,w-1,q-L), c-L]`.
- For fixed `S`, distributing spaces among `w-1` internal gaps (≥1) plus leading/trailing (≥0) gives `C(S+1, w)` (stars and bars).
- Sum spacing ways over valid `S`: `space_ways = Σ C(S+1, w)`.
- Answer = `2^A * space_ways - 1` (exclude original).
- Use big integers. Python has them built-in; C++ uses a custom `bigint`.
