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;
vector<int> V;
long long n, i, s = 0, d1, d2, hi, lo, mid, dm, w, m, k, c = 0;
long long digit(long long a) {
  long long p = 1, s = 0;
  if (!a) return 0;
  for (int i = 1; i < 18; i++) {
    p *= 10;
    if (a >= (p - 1)) {
      s += 9 * (p / 10) * i;
    } else {
      s += (a - (p / 10) + 1) * i;
      return s;
    }
  }
}
int main() {
  cin >> w >> m >> k;
  w = w / k;
  hi = 100000000000000000;
  lo = m;
  mid = (hi + lo) / 2;
  dm = digit(m - 1);
  while ((hi - lo) > 5) {
    d1 = digit(mid) - dm;
    d2 = digit(mid + 1) - dm;
    if (d1 <= w && d2 > w) {
      c = 1;
      break;
    } else if (d1 > w)
      hi = mid;
    else
      lo = mid;
    mid = (hi + lo) / 2;
  }
  if (c)
    cout << mid - m + 1;
  else {
    long long mx = 0;
    for (i = lo; i <= hi; i++) {
      d1 = digit(i) - dm;
      if (d1 <= w) {
        mx = i - m + 1;
      } else
        break;
    }
    cout << mx;
  }
}
