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 pow(long long n) {
  long long sum = 1;
  long long m = 10;
  while (n) {
    if (n % 2) sum *= m;
    m *= m;
    n /= 2;
  }
  return sum;
}
long long get(char s[]) {
  long long sum = 0;
  long long len = strlen(s);
  for (int i = len - 1; i >= 0; i--) {
    sum += (long long)(s[i] - '0') * pow(len - i - 1);
  }
  return sum;
}
int main() {
  long long w, mm, k;
  char m[30];
  cin >> w >> m >> k;
  w = w / k;
  if (!w) {
    cout << "0" << endl;
    return 0;
  }
  long long len = strlen(m);
  mm = get(m);
  long long sum = 0, sump = 0, ans = 0;
  while (sum <= w) {
    sum += len * (pow(len) - mm);
    if (sum > w) break;
    ans += pow(len) - mm;
    mm = pow(len);
    len++;
    sump = sum;
  }
  sump = w - sump;
  ans += sump / len;
  if (ans > 0)
    cout << ans << endl;
  else
    cout << "0" << endl;
  return 0;
}
