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 nd(long long n) {
  int cnt = 0;
  while (n) {
    cnt++;
    n /= 10;
  }
  return cnt;
}
long long mpow(long long b, int p) {
  long long res = 1;
  for (int i = 0; i < p; i++) res *= b;
  return res;
}
int main() {
  long long w, m, k;
  cin >> w >> m >> k;
  w = w / k;
  long long d = nd(m);
  long long base = mpow(10, d);
  long long cnt = 0;
  while (w >= (base - m) * d) {
    cnt += base - m;
    w -= (base - m) * d;
    m = base;
    base *= 10;
    d++;
  }
  long long l = 0, r = m * 9 + 1;
  d = nd(m);
  while (r - l > 1) {
    long long mid = (r + l) / 2;
    if (mid * d <= w && mid == (mid * d) / d)
      l = mid;
    else
      r = mid;
  }
  cout << cnt + l << endl;
  return 0;
}
