## 1. Abridged problem statement

There is a number hidden in the range `1..q`. You may repeatedly guess integers. Normally, after each guess you are told whether the hidden number is smaller, greater, or equal to your guess.

However, this machine answers with delay `c`: after your `i`-th guess, it gives the answer to your `(i-c)`-th guess. For the first `c` guesses, it outputs nothing.

Given `q` and `c`, find the minimum number of guesses needed in the worst case to guarantee finding the number.

Constraints:

```text
1 <= q <= 10^15
0 <= c <= 10^6
```

---

## 2. Detailed editorial

Let:

```text
f(n, c) = maximum size of an interval that can always be solved in at most n guesses
          when answers are delayed by c guesses.
```

We need the smallest `n` such that:

```text
f(n, c) >= q
```

For simplicity, write `f(n)` for fixed `c`.

### Base case

If `n <= c`, then no answer can arrive yet, because every answer is delayed by `c` guesses.

So:

```text
f(n) = 0 for n <= c
```

This also explains sample 2:

```text
q = 1, c = 1
```

Even if we guess `1` immediately, we only receive `"Correct"` after one more guess, so the answer is `2`.

---

### Recurrence

Consider an optimal strategy using `n` guesses.

Let the first guess be a splitter `m`.

Before the answer to the first guess arrives, we must already make `c` more guesses blindly. That means guesses:

```text
q_1, q_2, ..., q_{c+1}
```

are all fixed before we know whether the hidden number is smaller or greater than `q_1`.

The optimal strategy is to put those `c` blind follow-up guesses on one side of the first splitter.

Suppose we put them below the first guess.

Then there are three cases:

---

### Case 1: hidden number equals the first guess

Then we eventually receive `"Correct"`.

This accounts for `1` possible number.

---

### Case 2: hidden number is smaller than the first guess

The guesses `q_2, ..., q_n` form a delayed-search strategy of length `n - 1`.

So the lower side can contain at most:

```text
f(n - 1)
```

numbers.

---

### Case 3: hidden number is greater than the first guess

The `c` blind guesses after the first one were all below the splitter, so they are useless for this branch.

Only guesses:

```text
q_{c+2}, q_{c+3}, ..., q_n
```

remain useful.

There are:

```text
n - c - 1
```

such guesses.

So the upper side can contain at most:

```text
f(n - c - 1)
```

numbers.

---

Therefore:

```text
f(n) = f(n - 1) + f(n - c - 1) + 1
```

with:

```text
f(n) = 0 for n <= c
```

For `c = 0`, this becomes:

```text
f(n) = 2 * f(n - 1) + 1
```

so:

```text
f(n) = 2^n - 1
```

which is exactly ordinary binary search.

---

### Why this recurrence is optimal

The first guess splits the interval into lower part, the guessed value itself, and upper part.

Before the first answer arrives, the next `c` guesses are made without knowing which side is correct.

If some of these blind guesses are spent on the lower side, they may help only in the lower branch. If they are spent on the upper side, they may help only in the upper branch.

The best total coverage is achieved by committing all blind guesses to one side. Then one branch gets an almost full strategy of length `n - 1`, and the other branch loses those `c` guesses and gets only `n - c - 1` useful guesses.

Thus the maximum guaranteed searchable interval satisfies:

```text
f(n) = f(n - 1) + f(n - c - 1) + 1
```

---

### Computing the answer

We iterate `n = 1, 2, 3, ...` and compute `f(n)` until:

```text
f(n) >= q
```

Then `n` is the answer.

The recurrence only needs:

```text
f(n - 1)
f(n - c - 1)
```

So we do not need to store all values. A circular buffer of size `c + 2` is enough.

Why `c + 2`?

The largest backward distance needed is from `n` to `n - c - 1`, which is `c + 1`. To avoid overwriting needed values, we use size `c + 2`.

---

### Complexity

Let `answer` be the required number of guesses.

```text
Time complexity:  O(answer)
Memory complexity: O(c)
```

This is fast enough for the given constraints.

---

## 3. Provided C++ solution with detailed comments

```cpp
#include <bits/stdc++.h> 
// Includes almost all standard C++ library headers.

using namespace std;
// Allows using standard library names like vector, cin, cout without std:: prefix.

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    // Overloads output operator for pairs.
    // This is not actually needed in this solution, but is part of the author's template.
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    // Overloads input operator for pairs.
    // Also unused here.
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    // Overloads input operator for vectors.
    // Reads all elements of vector a.
    for(auto& x: a) {
        // Iterate by reference over every element.
        in >> x;
        // Read one element.
    }
    return in;
    // Return the input stream to allow chained input.
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    // Overloads output operator for vectors.
    // Prints all elements separated by spaces.
    for(auto x: a) {
        // Iterate over every element.
        out << x << ' ';
        // Print the element followed by a space.
    }
    return out;
    // Return the output stream to allow chained output.
};

int64_t q;
// q is the size of the search range, up to 10^15.
// int64_t is used because int would overflow.

int c;
// c is the delay parameter, up to 10^6.

void read() {
    // Reads input.
    cin >> q >> c;
}

void solve() {
    /*
        Let dp[n] = maximum number of possible hidden values that can be
        guaranteed to be found in n guesses with delay c.

        Base:
            dp[n] = 0 for n <= c

        Recurrence:
            dp[n] = dp[n - 1] + dp[n - c - 1] + 1

        Explanation:
            First guess splits the interval.
            One branch can use the remaining n - 1 guesses.
            The other branch loses the c blind guesses, so it gets n - c - 1 guesses.
            The first guessed value itself contributes 1.
    */

    int sz = c + 2;
    // We only need dp[n - 1] and dp[n - c - 1].
    // A circular buffer of size c + 2 is enough.

    vector<int64_t> dp(sz, 0);
    // Circular buffer initialized with zeros.
    // This also correctly represents dp[n] = 0 for n <= c.

    for(int64_t n = 1;; n++) {
        // Try increasing numbers of guesses n until the covered range reaches q.

        int64_t val = 0;
        // val will store dp[n].
        // By default it is 0, which is correct for n <= c.

        if(n > c) {
            // Only when n > c can we receive at least one answer.

            val = dp[(n - 1) % sz] + dp[(n - c - 1) % sz] + 1;
            // Apply recurrence:
            // dp[n] = dp[n - 1] + dp[n - c - 1] + 1.
            //
            // Values are accessed modulo sz because dp is a circular buffer.
        }

        dp[n % sz] = val;
        // Store dp[n] into the circular buffer.

        if(val >= q) {
            // If n guesses can cover at least q possible numbers,
            // then n is sufficient and is minimal because we iterate upward.

            cout << n << '\n';
            // Output the answer.

            return;
            // End solve().
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    // Speeds up C++ input/output by disabling synchronization with C stdio.

    cin.tie(nullptr);
    // Unties cin from cout for faster input.

    int T = 1;
    // The problem has exactly one test case.

    // cin >> T;
    // This line is commented out because there is no test-case count in input.

    for(int test = 1; test <= T; test++) {
        // Run once.

        read();
        // Read q and c.

        solve();
        // Compute and print the answer.
    }

    return 0;
    // Successful program termination.
}
```

---

## 4. Python solution with detailed comments

```python
import sys
# Import sys for fast input reading.


def solve():
    # Read all input tokens.
    data = sys.stdin.read().split()

    # q is the number of possible hidden values.
    q = int(data[0])

    # c is the response delay.
    c = int(data[1])

    # We define:
    #
    # dp[n] = maximum interval size solvable in n guesses.
    #
    # Base:
    #   dp[n] = 0 for n <= c
    #
    # Recurrence:
    #   dp[n] = dp[n - 1] + dp[n - c - 1] + 1
    #
    # We only need two older values:
    #   dp[n - 1]
    #   dp[n - c - 1]
    #
    # Therefore a circular buffer of size c + 2 is enough.
    size = c + 2

    # Initialize circular buffer with zeros.
    # These zeros also represent all base values dp[n] = 0 for n <= c.
    dp = [0] * size

    # Start checking n = 1, 2, 3, ...
    n = 1

    while True:
        # By default, dp[n] is 0.
        # This is correct when n <= c because no answer has arrived yet.
        value = 0

        if n > c:
            # Apply recurrence:
            #
            # dp[n] = dp[n - 1] + dp[n - c - 1] + 1
            #
            # Use modulo indexing because dp is stored as a circular buffer.
            value = dp[(n - 1) % size] + dp[(n - c - 1) % size] + 1

        # Store dp[n] in the circular buffer.
        dp[n % size] = value

        # Once we can cover at least q numbers, n guesses are sufficient.
        # Because n increases one by one, this n is minimal.
        if value >= q:
            print(n)
            return

        # Try the next number of guesses.
        n += 1


if __name__ == "__main__":
    # Run the solver.
    solve()
```

---

## 5. Compressed editorial

Let `f(n)` be the maximum number of possible hidden values solvable in `n` guesses with delay `c`.

If `n <= c`, no answer can arrive, so:

```text
f(n) = 0
```

For `n > c`, make the first guess as a splitter. The next `c` guesses are made blindly before the first answer arrives.

Put those blind guesses on one side. Then:

- one branch can use `n - 1` remaining guesses;
- the other branch loses the `c` blind guesses and can use only `n - c - 1` guesses;
- the first guessed value itself adds `1`.

Therefore:

```text
f(n) = f(n - 1) + f(n - c - 1) + 1
```

For `c = 0`, this becomes ordinary binary search:

```text
f(n) = 2f(n - 1) + 1
```

Compute `f(n)` increasing `n` until `f(n) >= q`. Since the recurrence only needs `f(n - 1)` and `f(n - c - 1)`, store values in a circular buffer of size `c + 2`.

Complexities:

```text
Time:   O(answer)
Memory: O(c)
```