#include <bits/stdc++.h>
using namespace std;
long long POW(int n) {
  long long mul = 1;
  while (n--) mul *= 10;
  return mul;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long w, m, k, sum = 0, n, cost, digit;
  while (cin >> w >> m >> k) {
    sum = 0;
    while (w > 0) {
      digit = log10(m) + 1;
      n = POW(digit) - m;
      cost = digit * k;
      if (n <= w / cost)
        w -= n * cost, sum += n, m += n;
      else
        sum += w / cost, w = 0;
    }
    cout << sum << endl;
  }
  return 0;
}

### Prompt

Generate 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;
long long POW(int n) {
  long long mul = 1;
  while (n--) mul *= 10;
  return mul;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long w, m, k, sum = 0, n, cost, digit;
  while (cin >> w >> m >> k) {
    sum = 0;
    while (w > 0) {
      digit = log10(m) + 1;
      n = POW(digit) - m;
      cost = digit * k;
      if (n <= w / cost)
        w -= n * cost, sum += n, m += n;
      else
        sum += w / cost, w = 0;
    }
    cout << sum << endl;
  }
  return 0;
}

```