#include <bits/stdc++.h>
using namespace std;
long long int getSum(long long int n) {
  long long int x = 1, sm = 0;
  long long int dg = 1;
  for (; x <= n; x *= 10, dg++) {
    sm += (min(x * 10, n + 1) - x) * dg;
  }
  return sm;
}
int main() {
  long long int w, m, k;
  cin >> w >> m >> k;
  long long int l = 0, r = 10000000000000000LL;
  while (l < r) {
    long long int mid = (l + r + 1) / 2;
    long long int ss = getSum(m + mid - 1) - getSum(m - 1);
    if (ss > w / k) {
      r = mid - 1;
    } else
      l = mid;
  }
  cout << l << 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;
long long int getSum(long long int n) {
  long long int x = 1, sm = 0;
  long long int dg = 1;
  for (; x <= n; x *= 10, dg++) {
    sm += (min(x * 10, n + 1) - x) * dg;
  }
  return sm;
}
int main() {
  long long int w, m, k;
  cin >> w >> m >> k;
  long long int l = 0, r = 10000000000000000LL;
  while (l < r) {
    long long int mid = (l + r + 1) / 2;
    long long int ss = getSum(m + mid - 1) - getSum(m - 1);
    if (ss > w / k) {
      r = mid - 1;
    } else
      l = mid;
  }
  cout << l << endl;
  return 0;
}

```