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;
int S(long long a) {
  int d = 1;
  while (a > 9) a /= 10, d++;
  return d;
}
long long get(long long a, long long p = 1) {
  if (a < 10) return a;
  long long m = 1;
  int d = 1;
  long long res = 0;
  while (a >= m * 10) {
    res += 9 * d * m;
    m *= 10;
    d += 1;
  }
  return res + (a - m + 1) * d;
}
bool ok(long long a, long long b) { return double(a) * b < LLONG_MAX; }
int main(void) {
  long long w, m, k;
  long long lb, mb, rb, res;
  cin >> w >> m >> k;
  res = -1;
  lb = 0;
  rb = 10000000000000000ll;
  while (lb <= rb) {
    mb = (lb + rb) / 2;
    if (ok((get(m + mb) - get(m - 1)), k) &&
        (get(m + mb) - get(m - 1)) * k <= w) {
      lb = mb + 1;
      res = max(res, mb);
    } else {
      rb = mb - 1;
    }
  }
  cout << rb + 1 << endl;
  return 0;
}
