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>
long long int seqSum(long long int x) {
  long long int sum = 0;
  long long int pow_of_ten = 1;
  long long int idx = 1;
  while (pow_of_ten * 10 <= x) {
    sum += pow_of_ten * 9 * idx;
    pow_of_ten *= 10;
    idx++;
  }
  sum += (x - pow_of_ten + 1) * idx;
  return sum;
}
long long int sequenceLength(long long int w, long long m) {
  long long int start = m - 1;
  long long int end = 100000000000000000;
  while (start != end) {
    long long int middle = (start + end) / 2 + 1;
    if (seqSum(middle) - seqSum(m - 1) > w) {
      end = middle - 1;
    } else {
      start = middle;
    }
  }
  return (start - m + 1);
}
int main() {
  long long int w, m, k;
  scanf("%lld %lld %lld", &w, &m, &k);
  w = w / k;
  long long int n = sequenceLength(w, m);
  printf("%lld", n);
  return 0;
}
