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 nextp10(long long m) {
  long long res = 1;
  while (res <= m) res *= 10LL;
  return res;
}
long long len(long long n) {
  if (n == 0) return 1;
  int res = 0;
  while (n > 0) {
    ++res;
    n /= 10;
  }
  return res;
}
int main() {
  long long w, m, k;
  while (cin >> w >> m >> k) {
    w /= k;
    --m;
    long long savem = m;
    while (w > 0) {
      long long p10 = nextp10(m) - 1;
      if (m == 0) p10 = 9;
      long long l = len(m);
      if (m == 0 || m < p10) {
        long long need = l * (p10 - m);
        if (w >= need) {
          m = p10;
          w -= need;
        } else {
          long long can = w / l;
          m = m + can;
          w = 0;
        }
      } else {
        long long need = l + 1;
        if (w >= need) {
          w -= need;
          ++m;
        } else {
          w = 0;
          break;
        }
      }
    }
    cout << m - savem << endl;
  }
  return 0;
}
