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>
long long n, m, k;
inline long long len(long long v) {
  long long s = 0;
  while (v) {
    v /= 10, s++;
  }
  return s;
}
long long cost(long long a, long long b) {
  if (a > b) return 0;
  if (a == b) return len(a);
  if (len(a) == len(b)) return ((b - a + 1) * len(a));
  long long mid = (a + b) / 2;
  return cost(a, mid) + cost(mid + 1, b);
}
bool decision(long long b) { return cost(m, m + b) <= n / k; }
inline long long f(long long v) {
  long long mid, lb = 0, ub = 20000000000000000LL;
  while (lb < ub) {
    mid = (lb + ub + 1) / 2;
    if (decision(mid))
      lb = mid;
    else
      ub = mid - 1;
  }
  if (ub == 0 && len(m) > n / k) return -1;
  return ub;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &k);
  printf("%I64d", f(m) + 1);
}
