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 MAXD = 18;
long long pot10[MAXD], fixed_digits[MAXD], pre_digits[MAXD];
int getNumDigits(long long n) {
  int res = 0;
  while (n) {
    n /= 10;
    res++;
  }
  return res;
}
long long getUpTo(long long n) {
  if (n == 0) return 0;
  int d = getNumDigits(n);
  long long res = pre_digits[d - 1] + (n - pot10[d - 1] + 1) * d;
  return res;
}
bool isValid(long long n, long long extra, long long money) {
  long long cost = getUpTo(n) - extra;
  return cost <= money;
}
int main() {
  long long w, m, k;
  cin >> w >> m >> k;
  pot10[0] = 1;
  pre_digits[0] = 0;
  for (int i = 1; i < MAXD; i++) {
    pot10[i] = pot10[i - 1] * 10;
    fixed_digits[i] = 9 * pot10[i - 1] * i;
    pre_digits[i] = pre_digits[i - 1] + fixed_digits[i];
  }
  long long less_than_m = getUpTo(m - 1);
  long long lo = m;
  long long hi = 1e17;
  long long threshold = w / k;
  while (hi - lo > 1) {
    long long mid = (hi + lo) / 2;
    if (isValid(mid, less_than_m, threshold)) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  if (isValid(lo, less_than_m, threshold))
    cout << lo - m + 1 << endl;
  else
    cout << 0 << endl;
  return 0;
}
