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;
int kol(long long x) {
  int i = 0;
  while (x) {
    x /= 10;
    i++;
  }
  return i;
}
long long power(long long x, int n) {
  long long base = x;
  for (int i = 0; i < n - 1; i++) x *= base;
  return x;
}
int main() {
  long long w, m, k, l, r, mid;
  cin >> w >> m >> k;
  l = 0;
  r = w / k;
  while (l != r) {
    mid = ceil(double(l + r) / 2);
    long long v = w;
    for (int i = kol(m); i <= kol(m + mid - 1); i++) {
      if (i == kol(m))
        if (i == kol(m + mid - 1))
          v -= mid * k * i;
        else
          v -= (pow(10, i) - m) * k * i;
      else if (i == kol(m + mid - 1))
        v -= ((m + mid) - pow(10, i - 1)) * k * i;
      else
        v -= power(10, i - 1) * 9 * k * i;
    }
    if (v >= 0)
      l = mid;
    else
      r = mid - 1;
  }
  cout << l;
  return 0;
}
