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>
unsigned long long sn(unsigned long long n) {
  unsigned long long count = (unsigned long long)0;
  while (n) {
    count += (unsigned long long)1;
    n /= (unsigned long long)10;
  }
  return count;
}
unsigned long long pow(unsigned long long d) {
  unsigned long long i, power = (unsigned long long)1;
  for (i = (unsigned long long)0; i < d; i += (unsigned long long)1) {
    power *= (unsigned long long)10;
  }
  return power;
}
int main() {
  unsigned long long w, m, k, n, length, cost, remaining, d;
  length = cost = (unsigned long long)0;
  scanf("%llu %llu %llu", &w, &m, &k);
  n = m;
  d = sn(n);
  while (1) {
    remaining = pow(d) - n;
    if (remaining * d * k + cost > w) {
      length += ((w - cost) / (d * k));
      break;
    } else {
      cost += (remaining * d * k);
      length += remaining;
    }
    n = pow(d);
    d += (unsigned long long)1;
  }
  printf("%llu", length);
  return 0;
}
