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;
unsigned long long diff(string a, string b) {
  if (b.length() > a.length()) return 0;
  string bpad(a);
  int x = b.length() - 1;
  for (int i = a.length() - 1; i >= 0; --i) {
    if (x < 0)
      bpad[i] = '0';
    else {
      bpad[i] = b[x];
      x--;
    }
  }
  unsigned long long ans = 0;
  int carry = 0;
  for (int i = 0; i < a.length(); ++i) {
    if (bpad[i] > a[i]) {
      ans--;
      ans *= 10;
      ans += ((unsigned long long)(10 + (a[i] - '0') - (bpad[i] - '0')));
    } else {
      ans *= 10;
      ans += ((unsigned long long)(a[i] - bpad[i]));
    }
  }
  return ans;
}
int main() {
  unsigned long long w, k;
  string m;
  cin >> w;
  cin >> m;
  cin >> k;
  unsigned long long ans = 0;
  string xxx(m.length(), '9');
  unsigned long long d = diff(xxx, m) + 1;
  while (d * k * m.length() <= w) {
    w -= d * k * m.length();
    string xm(m.length() + 1, '0');
    m = xm;
    m[0] = '1';
    string xxx(xm.length(), '9');
    ans += d;
    d = diff(xxx, m) + 1;
  }
  ans += (w) / (m.length() * k);
  cout << ans << '\n';
  return 0;
}
