#include <bits/stdc++.h>
using namespace std;
int S(long long m) {
  int cnt = 0;
  while (m) {
    cnt++;
    m /= 10;
  }
  return cnt;
}
int main() {
  long long w, m, k;
  cin >> w >> m >> k;
  int ok = 1;
  long long ans = 0;
  while (ok) {
    int cnt = S(m);
    long long b = 1;
    for (int i = 0; i < cnt; ++i) b *= 10;
    b--;
    long long rest = b - m + 1;
    if (rest <= w / (cnt * k)) {
      w -= cnt * k * rest;
      ans += rest;
      m += rest;
    } else {
      ok = 0;
      ans += w / (cnt * k);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 S(long long m) {
  int cnt = 0;
  while (m) {
    cnt++;
    m /= 10;
  }
  return cnt;
}
int main() {
  long long w, m, k;
  cin >> w >> m >> k;
  int ok = 1;
  long long ans = 0;
  while (ok) {
    int cnt = S(m);
    long long b = 1;
    for (int i = 0; i < cnt; ++i) b *= 10;
    b--;
    long long rest = b - m + 1;
    if (rest <= w / (cnt * k)) {
      w -= cnt * k * rest;
      ans += rest;
      m += rest;
    } else {
      ok = 0;
      ans += w / (cnt * k);
    }
  }
  cout << ans << endl;
  return 0;
}

```