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 a[20];
unsigned long long w, k, m, cnt = 0;
const unsigned long long maxN = 1e16 + 10;
inline unsigned long long mul(unsigned long long a, unsigned long long b) {
  if ((maxN + b - 1) / b > a)
    return a * b;
  else
    return maxN;
}
unsigned long long Cost(unsigned long long x) {
  unsigned long long now = 19, tot = 0;
  while (now >= cnt) {
    if (a[now] <= x) {
      tot += (x - a[now] + 1) * now;
      x = a[now] - 1;
    }
    now--;
  }
  return tot;
}
unsigned long long BS() {
  a[2] = 10;
  for (int(i) = (3); (i) < (int)(20); ++(i)) a[i] = a[i - 1] * 10;
  while (cnt < 20 and a[cnt] <= m) a[cnt] = m, cnt++;
  cnt--;
  unsigned long long fi = m, se = maxN * 100;
  while (fi < se) {
    unsigned long long mid = (fi + se + 1) / 2;
    if (mul(Cost(mid), k) > w)
      se = mid - 1;
    else
      fi = mid;
  }
  if (mul(Cost(fi), k) > w) fi--;
  return fi;
}
int main() {
  scanf("%I64d%I64d%I64d", &w, &m, &k);
  printf("%I64d\n", (BS() - m + 1));
  return 0;
}
