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 m, k, w;
long long len(long long a) {
  long long ret = 0;
  if (a == 0) return 1;
  while (a != 0) {
    a /= 10;
    ret++;
  }
  return ret;
}
bool judge(long long n) {
  long long sum = 0, weg = 1, leng = 0, i;
  long long tmp = m;
  leng = len(tmp);
  for (i = 0; i < leng; i++) weg *= 10;
  while (n > 0) {
    if (n > weg - tmp) {
      n -= (weg - tmp);
      sum += (weg - tmp) * leng;
    } else {
      sum += n * leng;
      break;
    }
    tmp = weg;
    weg *= 10;
    leng++;
  }
  if (sum <= w)
    return true;
  else
    return false;
}
int main() {
  long long l, r, mid, ans;
  while (~scanf("%I64d%I64d%I64d", &w, &m, &k)) {
    w /= k;
    l = 0;
    r = w / len(m);
    ans = 0;
    judge(100);
    while (l <= r) {
      mid = (l + r) / 2;
      if (judge(mid)) {
        l = mid + 1;
        ans = mid;
      } else {
        r = mid - 1;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
