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
#include <bits/stdc++.h>
using namespace std;
long long arr[] = {0,
                   9,
                   99,
                   999,
                   9999,
                   99999,
                   999999,
                   9999999,
                   99999999,
                   999999999,
                   9999999999,
                   99999999999,
                   999999999999,
                   9999999999999,
                   99999999999999,
                   999999999999999,
                   9999999999999999,
                   99999999999999999};
int main() {
  long long w, m, k;
  scanf("%I64d%I64d%I64d", &w, &m, &k);
  w /= k;
  int start;
  for (int i = 1; i <= 18; i++) {
    if (arr[i] >= m) {
      start = i;
      break;
    }
  }
  long long sum = 0;
  for (int i = start; i <= 18; i++) {
    if (w >= (arr[i] - m + 1) * i) {
      sum += arr[i] - m + 1;
      w -= (arr[i] - m + 1) * i;
      m += (arr[i] - m + 1);
    } else {
      sum += w / i;
      break;
    }
  }
  printf("%I64d\n", sum);
}
