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 ten[50];
long long g(long long n) {
  long long ans = 0;
  for (long long i = 1; i <= n - 1; i++) ans += i * 9 * ten[i - 1];
  return ans;
}
long long f(long long n) {
  if (n == 0) return 0;
  long long deg = 10;
  long long x = 1;
  while (deg <= n) {
    deg *= 10;
    x++;
  }
  deg /= 10;
  return g(x) + x * (n - deg + 1);
}
long long w, m, k;
int main() {
  ten[0] = 1;
  for (long long i = 1; i < 19; i++) ten[i] = (long long)10 * ten[i - 1];
  scanf("%I64d %I64d %I64d", &w, &m, &k);
  w /= k;
  long long l = 0, r = w;
  while (l < r) {
    long long s = (l + r + 1) / 2;
    if (f(m + s - 1) - f(m - 1) <= w)
      l = s;
    else
      r = s - 1;
  }
  printf("%I64d", l);
}
