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 ans = 0;
long long w, m, k;
long long n;
const long long powConv[] = {0,
                             1,
                             10,
                             100,
                             1000,
                             10000,
                             100000,
                             1000000,
                             10000000,
                             100000000,
                             1000000000,
                             10000000000,
                             100000000000,
                             1000000000000,
                             10000000000000,
                             100000000000000,
                             1000000000000000,
                             10000000000000000,
                             100000000000000000,
                             1000000000000000000};
long long S(long long a) {
  if (a < 10) return 1;
  return 1 + S(a / 10);
}
void getAns() {
  long long s1 = S(m);
  if (m != powConv[s1]) {
    if ((powConv[s1 + 1] - m) * s1 < n) {
      ans += powConv[s1 + 1] - m;
      n -= (powConv[s1 + 1] - m) * s1;
      s1++;
      m = powConv[s1];
    }
  }
  while ((powConv[s1 + 1] - powConv[s1]) * s1 < n) {
    ans += (powConv[s1 + 1] - powConv[s1]);
    n -= (powConv[s1 + 1] - powConv[s1]) * s1;
    s1++;
    m = powConv[s1];
  }
  long long s2 = n / s1;
  long long more = s2;
  ans += more;
  m += more;
}
int main() {
  cin >> w >> m >> k;
  n = w / k;
  getAns();
  cout << ans;
  return 0;
}
