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 Mul[21], w, m, k, ans = 0;
void predo() {
  Mul[0] = 1;
  for (unsigned long long i = 1; i <= 17; i++) Mul[i] = Mul[i - 1] * 10;
}
int main() {
  scanf("%I64u %I64u %I64u", &w, &m, &k);
  predo();
  w /= k;
  unsigned long long tmp = m, l = 0;
  while (tmp) tmp /= 10, ++l;
  unsigned long long res = Mul[l] - m;
  if (w <= res * l)
    printf("%I64u\n", w / l);
  else {
    w -= res * l;
    ans += res;
    for (unsigned long long i = l + 1; i <= 17; i++) {
      unsigned long long tmp = Mul[i] - Mul[i - 1];
      if (w <= tmp * i) {
        ans += w / i;
        printf("%I64u\n", ans);
        break;
      } else {
        w -= tmp * i;
        ans += tmp;
      }
    }
  }
  return 0;
}
