## 1) Abridged problem statement

Given positive integers **A** (up to 3000 decimal digits) and **B** (1 ≤ B ≤ 10⁶), represent **A** as a sum of signed powers of **B**:

\[
A=\sum_{i=1}^{n} s_i \, B^{k_i}, \quad s_i\in\{-1,+1\},\ k_i\in\mathbb{Z}
\]

Minimize **n** (the number of terms). Output this minimum **n**.

---

## 2) Detailed editorial (how the solution works)

### Key observation 1: combine equal exponents ⇒ “digits” in base B
If the same exponent appears multiple times, we can combine:
- \(+B^k + B^k = 2B^k\)
- \(+B^k - B^k = 0\)

So any solution corresponds to choosing integer coefficients \(c_k\) such that:
\[
A=\sum_k c_k B^k
\]
and each coefficient \(c_k\) is an integer, where writing \(c_k\) as a sum of \(\pm 1\) needs \(|c_k|\) terms. Therefore:
\[
n = \sum_k |c_k|
\]
So we want a base‑\(B\) expansion with (possibly negative) “digits” minimizing sum of absolute digit values.

### Key observation 2: negative exponents don’t help (for minimal n)
Allowing \(k_i < 0\) introduces fractions \(B^{-t}\). For the total sum to be an integer \(A\), all fractional parts must cancel perfectly. Any such cancellation requires at least as many \(\pm B^{-t}\) terms as simply “carrying” them upward into nonnegative powers. Thus an optimal solution exists using only \(k_i \ge 0\).  
So we only need an integer signed-digit representation.

### Special case: B = 1
All powers \(1^k = 1\). Then the expression becomes a sum of \(n\) many \(\pm 1\), and to reach positive integer A minimally, we just use A copies of +1.  
Answer: **n = A** (A is huge, but we can print it as a bigint).

### For B ≥ 2: dynamic programming on base-B digits with carry
Write A in standard base B:
\[
A = d_0 + d_1 B + d_2 B^2 + \cdots,\quad 0\le d_i < B
\]

We will convert it into a *balanced* form where each digit becomes either:
- \(r\) (nonnegative), or
- \(r-B\) (negative),
while pushing a carry to the next position.

Process from least significant digit upward with an incoming carry \(c \in \{0,1\}\).

At position i:
- Let \(t = d_i + c\).
- Let \(r = t \bmod B\), \(q = \lfloor t/B \rfloor\).

Two choices to represent this position:

1) **Use digit \(r\)**  
   Cost contributes \(|r| = r\) terms.  
   Next carry is \(q\) (0 or 1).

2) **Use digit \(r-B\)** (only if \(r \ne 0\))  
   This digit is negative, absolute value is \(|r-B|=B-r\).  
   It effectively adds one more carry to next digit, so next carry is \(q+1\).
   In this problem, it turns out carry never exceeds 1 in optimal transitions:
   - If \(q=1\), then \(r=0\), so branch 2 is impossible.
   - If \(r\ne 0\), then \(q=0\), so \(q+1=1\).

So carry states remain just {0,1}.

#### DP definition
Let `dp[i][c]` = minimal cost (minimal number of terms) to represent suffix starting at digit i, given incoming carry `c` into digit i.

Base (past the top digit, i = L):
- `dp[L][0] = 0` (nothing left)
- `dp[L][1] = 1` (a leftover carry 1 means we need one extra \(+B^L\) term)

Transition for i from L-1 down to 0:
- compute t, r, q as above
- option1 = r + dp[i+1][q]
- option2 = (B-r) + dp[i+1][q+1] if r != 0
- dp[i][c] = min(option1, option2)

Answer is `dp[0][0]`.

### Complexity
Let \(L\) be number of base‑B digits of A.
- Converting to base B: \(O(L)\) bigint divisions by int (implemented)
- DP: \(O(L)\)

Memory: O(L). Fits easily.

---

## 3) Provided C++ solution with detailed line-by-line comments

```cpp
#include <bits/stdc++.h>
using namespace std;

// Pretty-print a pair as "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Read a pair from stream
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Read a vector by reading each element
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x: a) in >> x;
    return in;
};

// Print a vector with spaces (not used in this problem)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x: a) out << x << ' ';
    return out;
};

// Big integer base: each "limb" stores 9 decimal digits (1e9)
const int base = 1000000000;
const int base_digits = 9;

// A signed big integer implementation supporting +, -, *, /, %, etc.
struct bigint {
    vector<int> z;  // limbs, least significant first
    int sign;       // +1 or -1

    bigint() : sign(1) {}
    bigint(long long v) { *this = v; }      // assign from 64-bit
    bigint(const string& s) { read(s); }    // parse from string

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

    // assignment from long long
    void operator=(long long v) {
        sign = 1;
        if (v < 0) sign = -1, v = -v;
        z.clear();
        // store in base 1e9
        for (; v > 0; v /= base) z.push_back(v % base);
    }

    // addition
    bigint operator+(const bigint& v) const {
        // same sign => add magnitudes
        if (sign == v.sign) {
            bigint res = v; // start from v, add this into it
            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;
        }
        // different signs => a + (-b) = a - b
        return *this - (-v);
    }

    // subtraction
    bigint operator-(const bigint& v) const {
        if (sign == v.sign) {
            // if |this| >= |v| we can subtract directly
            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;
            }
            // otherwise result is negative: a-b = -(b-a)
            return -(v - *this);
        }
        // different signs: a - (-b) = a + b
        return *this + (-v);
    }

    // multiply by int
    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] * 1LL * v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
        }
        trim();
    }

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

    // division with remainder for bigint / bigint (uses normalization)
    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);     // scaling factor
        bigint a = a1.abs() * norm;              // normalized numerator
        bigint b = b1.abs() * norm;              // normalized denominator
        bigint q, r;
        q.z.resize(a.z.size());                  // quotient has <= that many limbs

        // long division from most significant limb to least
        for (int i = (int)a.z.size() - 1; i >= 0; i--) {
            r *= base;                           // shift remainder by base
            r += a.z[i];                         // add next limb
            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(); // trial digit
            r -= b * d;
            while (r < 0) {                      // fix if trial too big
                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);           // un-normalize remainder
    }

    // integer square root (not used by this problem)
    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 = (int)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; }

    // divide by int
    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 * 1LL * base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

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

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

    // compound ops
    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; }

    // comparisons
    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 = (int)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; }

    // remove leading zero limbs and fix sign for zero
    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]); }

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

    // absolute value
    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign; // makes sign = +1
        return res;
    }

    // convert to long long (unsafe if too large; not used)
    long long longValue() const {
        long long res = 0;
        for (int i = (int)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;
    }

    // parse decimal string into bigint
    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 = (int)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;
    }

    // convert vector of digits from one decimal-digit grouping to another
    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;

    // Karatsuba multiplication helper (not really needed for this DP problem,
    // but included in the bigint implementation for speed)
    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = (int)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;
    }

    // full bigint multiplication using base conversion + Karatsuba
    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;
    }
};

bigint a; // huge A
int b;    // base B

// read input
void read() { cin >> a >> b; }

void solve() {
    // If B == 1: all powers are 1, so we need exactly A times +1 to sum to A.
    if (b == 1) {
        cout << a << '\n';
        return;
    }

    // Extract base-B digits of A (least significant first) using bigint % int and / int.
    vector<int> digits;
    bigint x = a;
    while (!x.isZero()) {
        digits.push_back(x % b);
        x /= b;
    }
    int L = (int)digits.size(); // number of base-B digits

    // dp[i][c] stored as array<2>, for carry c in {0,1}
    vector<array<int64_t, 2>> dp(L + 1);

    // Base case: beyond top digit
    dp[L][0] = 0; // no carry => no extra terms
    dp[L][1] = 1; // leftover carry => one extra +B^L term

    // Process digits from most significant down to least (i decreasing)
    for (int i = L - 1; i >= 0; i--) {
        for (int c = 0; c < 2; c++) {
            int t = digits[i] + c; // digit plus incoming carry
            int r = t % b;         // remainder (candidate digit if we don't borrow)
            int q = t / b;         // next carry (0 or 1)

            // Option 1: use digit r (cost r) and pass carry q
            int64_t best = (int64_t)r + dp[i + 1][q];

            // Option 2: use digit r - B (negative), cost B-r, pass carry q+1
            // Only meaningful if r != 0 (otherwise r-B would be -B, not allowed by scheme)
            if (r != 0) {
                best = min(best, (int64_t)(b - r) + dp[i + 1][q + 1]);
            }

            dp[i][c] = best;
        }
    }

    // Minimal number of +/- power terms
    cout << dp[0][0] << '\n';
}

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

    int T = 1; // single test case
    for (int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

## 4) Python solution (same algorithm) with detailed comments

```python
import sys

def solve() -> None:
    # Read A as Python int (supports arbitrarily large integers)
    A_line = sys.stdin.readline().strip()
    B_line = sys.stdin.readline().strip()

    A = int(A_line)
    B = int(B_line)

    # Special case: B = 1
    # All powers are 1, so we need exactly A copies of +1 to sum to A.
    if B == 1:
        print(A)
        return

    # Convert A to base-B digits, least significant first.
    digits = []
    x = A
    while x > 0:
        digits.append(x % B)
        x //= B
    L = len(digits)

    # dp_next[c] will store dp[i+1][c] while iterating i downward.
    # Base case at i = L:
    # dp[L][0] = 0, dp[L][1] = 1 (leftover carry becomes one extra term)
    dp_next0, dp_next1 = 0, 1

    # Process digits from most significant to least (i = L-1 ... 0)
    for i in range(L - 1, -1, -1):
        d = digits[i]

        # Compute dp[i][0] and dp[i][1]

        # For carry c=0:
        t = d
        r = t % B
        q = t // B  # always 0 here because 0 <= d < B
        best0 = r + (dp_next0 if q == 0 else dp_next1)
        if r != 0:
            # borrow: digit becomes r-B, next carry q+1 = 1
            best0 = min(best0, (B - r) + dp_next1)

        # For carry c=1:
        t = d + 1
        r = t % B
        q = t // B  # either 0 or 1
        best1 = r + (dp_next0 if q == 0 else dp_next1)
        if r != 0:
            # This is only possible when q == 0, hence q+1 == 1 is valid
            best1 = min(best1, (B - r) + dp_next1)

        # Shift window: dp_next becomes dp[i]
        dp_next0, dp_next1 = best0, best1

    # Answer is dp[0][0]
    print(dp_next0)

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

---

## 5) Compressed editorial

- Combine equal powers: minimizing term count equals minimizing \(\sum |c_k|\) in an integer expansion \(A=\sum c_k B^k\).
- An optimal solution uses only nonnegative exponents.
- For **B=1**, answer is **A**.
- For **B≥2**, write A in base B digits \(d_i\). Do DP from low to high using carry \(c\in\{0,1\}\).
  - At digit i: \(t=d_i+c\), \(r=t\bmod B\), \(q=\lfloor t/B\rfloor\).
  - Choose digit \(r\) (cost \(r\), next carry \(q\)) or digit \(r-B\) (cost \(B-r\), next carry \(q+1\), only if \(r\ne0\)).
- Base: dp[L][0]=0, dp[L][1]=1.
- Output dp[0][0].