#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int w, m, k;
  long long int sum = 0, mul = 10, count = 1, ans = 0;
  cin >> w >> m >> k;
  while (sum < w) {
    if (mul - m > 0) {
      if ((mul - m) > (float)(w - sum) / (float)(count * k)) {
        ans = ans + (w - sum) / (k * count);
        break;
      } else {
        sum += (mul - m) * count * k;
        ans += mul - m;
        count++;
        m = mul;
        mul *= 10;
      }
    } else {
      mul *= 10;
      count++;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

We'll define S(n) for positive integer n as follows: the number of the n's digits in the decimal base. For example, S(893) = 3, S(114514) = 6.

You want to make a consecutive integer sequence starting from number m (m, m + 1, ...). But you need to pay S(n)·k to add the number n to the sequence.

You can spend a cost up to w, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.

Input

The first line contains three integers w (1 ≤ w ≤ 1016), m (1 ≤ m ≤ 1016), k (1 ≤ k ≤ 109).

Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

The first line should contain a single integer — the answer to the problem.

Examples

Input

9 1 1


Output

9


Input

77 7 7


Output

7


Input

114 5 14


Output

6


Input

1 1 2


Output

0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int w, m, k;
  long long int sum = 0, mul = 10, count = 1, ans = 0;
  cin >> w >> m >> k;
  while (sum < w) {
    if (mul - m > 0) {
      if ((mul - m) > (float)(w - sum) / (float)(count * k)) {
        ans = ans + (w - sum) / (k * count);
        break;
      } else {
        sum += (mul - m) * count * k;
        ans += mul - m;
        count++;
        m = mul;
        mul *= 10;
      }
    } else {
      mul *= 10;
      count++;
    }
  }
  cout << ans << endl;
  return 0;
}

```