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 dp[20], tnu[20];
void init() {
  dp[0] = 0;
  dp[1] = 10;
  tnu[1] = 1;
  for (int i = 2; i < 19; i++)
    dp[i] = i * dp[i - 1] * 9 + dp[i - 1], tnu[i] = tnu[i - 1] * 10;
  tnu[0] = 0;
}
long long solve(long long w, long long m, int k) {
  int num = 0;
  long long temp = m;
  long long en;
  while (temp > 0) temp = temp / 10, num++;
  w = w / k;
  if (w < num) return 0;
  en = m;
  while (w > 0) {
    if ((tnu[num + 1] - en) < w / num) {
      w = w - (tnu[num + 1] - en) * num;
      num++;
      en = tnu[num];
    } else {
      en = en + w / (num)-1;
      w = 0;
    }
  }
  return en - m + 1;
}
int main() {
  long long w, m;
  int k;
  cin >> w >> m >> k;
  init();
  cout << solve(w, m, k);
  return 0;
}
