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 int pow10(int k) {
  long long int ans = 1;
  for (int i = 0; i < k; i++) {
    ans *= 10;
  }
  return ans;
}
int main() {
  long long int w = 0, m = 0;
  long long int k = 0;
  cin >> w >> m >> k;
  int digm = 0;
  long long int mc = m;
  while (mc > 0) {
    digm++;
    mc /= 10;
  }
  int minpow = digm;
  unsigned long long int sum = (pow10(digm) - m) * minpow * k;
  long long int ans = 0;
  long long int a = m;
  long long int lsum = 0;
  while (sum < w) {
    lsum = sum;
    ans += pow10(minpow) - a;
    a = pow10(minpow);
    minpow++;
    unsigned long long int temp =
        (pow10(minpow) - pow10(minpow - 1)) * minpow * k;
    sum += temp;
  }
  long long int rem = w - lsum;
  ans += rem / (k * minpow);
  cout << ans;
}
