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;
template <class T>
inline bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c<'0' | c> '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
const long long N = 100005;
long long ara[N];
int main() {
  unsigned long long w, m, k;
  cin >> w >> m >> k;
  unsigned long long val2 = m;
  unsigned long long digit = 0;
  while (val2 > 0) {
    val2 /= 10;
    ++digit;
  }
  unsigned long long curr = 1;
  for (long long i = 1; i <= digit; ++i) {
    curr *= 10;
  }
  unsigned long long ans = 0;
  unsigned long long val = curr - m;
  if (digit * k * val <= w) {
    ans += val;
    w -= (digit * k * val);
  } else {
    ans += w / (digit * k);
    return cout << ans, 0;
  }
  while (true) {
    unsigned long long prev = curr * 10;
    ++digit;
    unsigned long long val = prev - curr;
    if (digit * k * val <= w) {
      ans += val;
      w -= (digit * k * val);
    } else {
      ans += w / (digit * k);
      return cout << ans, 0;
    }
    curr = prev;
  }
}
