<|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.

295. Identifier Duplicated!
time limit per test: 0.25 sec.
memory limit per test: 65536 KB
input: standard
output: standard



On the first of April Qc cracked one of the most popular Karelian chats. Before his unlawful action only Latin letters, digits and space characters could be used in nicknames. But now users can use digits, spaces and both Latin and Russian letters. Certainly, this change brings the possibility of multing. Multing means that the new user is able to register under a nickname that looks like the existing one. For example, if there exists the user with the nickname "Hedgehog", the new user can get a nickname by replacing the first Latin letter "H" with Russian "H". Moreover, it is possible to change several letters at once; add or remove spaces.
Qc wants to know how many different mults exist for the given nickname.
Let us determine the set of Latin letters that are readable as Russian ones:
A, B, C, E, H, K, M, O, P, T, X, a, c, e, o, p, x, y.
Two words are mults of each other if one can be transformed to another by replacing one or several of its Latin letters with its Russian analog or vice versa, changing the number of spaces between words, adding or removing leading and trailing spaces. Note, that it is not allowed to delete all spaces between a pair of words, or add new spaces in the middle of the word.

Input
The first line of the input file contains two numbers q and c (1 <= q <= c <= 63) separated by one space. The length of registered nickname must be not less than q letters and no more than c. The second line contains nickname consisting only of Latin letters, digits and space characters.

Output
Output the number of mults for the given nickname under the given restrictions.

Sample test(s)

Input
3 15
Hedgehog

Output
575
Author:	Anton Golubev, Petrazavodsk SU
Resource:	Anton Golubev (Hedgehog)'s Contest #2 from Annual Summer Russian Teams Meeting in Petrozavodsk State University
Date:	August 26, 2005

<|response|>
## 1) Abridged problem statement

Given integers `q, c (1 ≤ q ≤ c ≤ 63)` and a nickname string consisting of **Latin letters, digits, and spaces**.

After a change, users may:
- Replace any occurrence of certain **ambiguous** Latin letters with their visually identical Russian counterparts (and vice versa). Ambiguous set:
  `A B C E H K M O P T X a c e o p x y` (18 letters)
- Change spacing: add/remove leading/trailing spaces and change the number of spaces between words, **but**
  - words must remain separated (cannot remove all spaces between two neighboring words),
  - cannot insert spaces inside a word.

Count how many *different* "mult" nicknames can be formed with total length between `q` and `c` inclusive. Do **not** count the original nickname itself.

The answer can be very large → use big integers.

---

## 2) Key observations needed to solve the problem

1) **Letter changes are independent of spaces.**
If the nickname contains `A` occurrences of ambiguous letters, each such character can be left as-is or swapped → `2^A` letter variants.

2) **Words are fixed; only spaces between/around words vary.**
Split the nickname into `w` words (maximal non-space substrings). Let:
- `L` = total number of non-space characters (sum of word lengths).
- `S` = total number of spaces in the final nickname.
Final length is `L + S`, so `q ≤ L + S ≤ c`.

3) **Spacing constraints become a stars-and-bars count.**
We distribute `S` spaces into `w+1` "gaps":
- leading gap (≥ 0),
- `w-1` internal gaps between consecutive words (each **≥ 1**),
- trailing gap (≥ 0).

For fixed `S`, the number of valid distributions is:
\[
\binom{S+1}{w}
\]
(derived by subtracting 1 from each internal gap to make all variables nonnegative).

4) **Valid range for `S`:**
- `S ≥ 0`
- `S ≥ w-1` (at least one space between each adjacent pair of words)
- `S ≥ q - L` (to reach minimum length)
- `S ≤ c - L` (maximum length)

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

If `S_min > S_max` → answer is `0`.

5) **Exclude the original nickname.**
The counting includes the original (no swaps + original spaces), so subtract 1 at the end.

---

## 3) Full solution approach based on the observations

1) Parse `q, c`, then read the whole nickname line (may contain spaces).
2) Split the nickname into words by spaces (ignore multiple spaces).
   - Let `w` be the number of words.
3) Compute:
   - `L`: total length of all words (non-space characters)
   - `A`: number of characters among those words belonging to the ambiguous set
4) Compute `letterWays = 2^A` using big integer arithmetic (the C++ code below uses a vendored `bigint`; Python ints are arbitrary precision).
5) Determine `S_min, S_max`. If invalid, print `0`.
6) Compute:
\[
spaceWays = \sum_{S=S_{min}}^{S_{max}} \binom{S+1}{w}
\]
7) Total variants (including original): `letterWays * spaceWays`.
8) Output `letterWays * spaceWays - 1`.

Complexity is tiny because `c ≤ 63`, so the sum has at most 64 terms and combinations are small-parameter (though the results are huge).

---

## 4) C++ implementation with detailed comments

```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;
}
```

---

## 5) Python implementation with detailed comments

```python
import sys
import math

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

    # Read nickname line (may contain spaces)
    s = sys.stdin.readline().rstrip("\n")

    # Ambiguous Latin letters
    amb = set("ABCEHKMOPTXaceopxy")

    # Split into words (collapse multiple spaces)
    words = [w for w in s.split(' ') if w != '']
    w = len(words)

    # L = total non-space length, A = ambiguous occurrences
    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 (Python int is arbitrary precision)
    letter_ways = 1 << A

    # Valid total spaces range
    S_min = max(0, w - 1, q - L)
    S_max = c - L
    if S_min > S_max:
        print(0)
        return

    # Spacing variants for fixed S: C(S+1, w)
    space_ways = 0
    for S in range(S_min, S_max + 1):
        space_ways += math.comb(S + 1, w)

    # Multiply independent choices and exclude original nickname
    ans = letter_ways * space_ways - 1
    print(ans)

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

These implementations follow directly from the combinatorial decomposition:
- independent letter flips (`2^A`)
- independent space redistributions (`Σ C(S+1, w)` over feasible lengths)
- minus one to exclude the original nickname.
