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;
const int maxn = 5e5 + 10;
const int inf = 0x3f3f3f3f;
const long long llinf = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1000000007;
long long a[20];
long long pow(long long x) {
  long long tem = 1;
  for (long long i = 1; i <= x; i++) {
    tem *= 10;
  }
  return tem;
}
int main() {
  long long w, m, k;
  scanf("%I64d%I64d%I64d", &w, &m, &k);
  for (long long i = 1; i <= 18; i++) {
    a[i] = pow(i);
  }
  w /= k;
  long long t = 0;
  long long tem = m;
  while (tem) {
    tem /= 10;
    t++;
  }
  long long ans = 0;
  for (long long i = t; i <= 20; i++) {
    if (i == t) {
      long long sum = a[i] - m;
      if (sum * i < w) {
        w -= sum * i;
        ans += sum;
      } else {
        ans += w / i;
        break;
      }
    } else {
      long long sum = a[i] - a[i - 1];
      if (sum * i < w) {
        w -= sum * i;
        ans += sum;
      } else {
        ans += w / (i);
        break;
      }
    }
  }
  printf("%I64d\n", ans);
}
