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;
const int N = 1000000;
long long int int_len(long long int n) {
  long long int res = 0;
  while (n) {
    n /= 10ll;
    ++res;
  }
  return res;
}
long long int bin_pow(long long int n) {
  long long int res = 1ll, a = 10ll;
  while (n) {
    if (n & 1ll) res *= a;
    a *= a;
    n >>= 1ll;
  }
  return res;
}
long long int p, n, k, cnt, ans;
int main() {
  cin >> p >> n >> k;
  cnt = bin_pow(int_len(n)) - n;
  if (p / k < 1.0 * cnt * int_len(n)) {
    cout << p / (k * int_len(n)) << "\n";
    return 0;
  }
  ans += cnt;
  p -= cnt * k * int_len(n);
  for (int i = 1;; ++i) {
    cnt = bin_pow(int_len(n) + i) - bin_pow(int_len(n) + i - 1ll);
    if (p / k < 1.0 * cnt * (int_len(n) + i)) {
      cout << ans + p / (k * (int_len(n) + i)) << "\n";
      return 0;
    } else {
      ans += cnt;
      p -= cnt * (int_len(n) + i) * k;
    }
  }
  return 0;
}
