#include <bits/stdc++.h>
using namespace std;
long long solve(long long a) {
  long long idx = 1;
  long long ans = 0;
  long long len = 1;
  while (1) {
    if (idx * 10 > a) {
      ans += (a - idx + 1) * len;
      break;
    }
    ans += idx * 9 * len;
    idx *= 10;
    len++;
  }
  return ans;
}
int main() {
  long long w, m, k;
  cin >> w >> m >> k;
  w = w / k;
  long long low = m - 1, high = 20000000000000000LL;
  long long mid;
  while (1) {
    mid = (low + high) / 2 + 1;
    if (solve(mid) - solve(m - 1) > w)
      high = mid - 1;
    else
      low = mid;
    if (low == high) {
      cout << low - m + 1;
      break;
    }
  }
}

### Prompt

Create a solution in CPP for 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;
long long solve(long long a) {
  long long idx = 1;
  long long ans = 0;
  long long len = 1;
  while (1) {
    if (idx * 10 > a) {
      ans += (a - idx + 1) * len;
      break;
    }
    ans += idx * 9 * len;
    idx *= 10;
    len++;
  }
  return ans;
}
int main() {
  long long w, m, k;
  cin >> w >> m >> k;
  w = w / k;
  long long low = m - 1, high = 20000000000000000LL;
  long long mid;
  while (1) {
    mid = (low + high) / 2 + 1;
    if (solve(mid) - solve(m - 1) > w)
      high = mid - 1;
    else
      low = mid;
    if (low == high) {
      cout << low - m + 1;
      break;
    }
  }
}

```