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>
#pragma comment(linker, "/STACK:65777216")
const long long INF = 1000000000;
const double PI = 3.14159265358979323846;
const long long MOD = 1000000000 + 7;
using namespace std;
long long pow(long long q) {
  long long t = 1;
  for (int i(0); i < q; i++) {
    t *= 10;
  }
  return t;
}
int main() {
  long long w, m, k, ans(0);
  cin >> w >> m >> k;
  k = w / k;
  long long t = m, n = 0;
  while (t) {
    n++;
    t /= 10;
  }
  if (m == 0) n = 1;
  t = n;
  if (k - (pow(t) - m) * t > 0) {
    ans += (pow(t) - m);
    k -= (pow(t) - m) * t;
  } else {
    ans += (k / t);
    k = 0;
  }
  while (k) {
    t++;
    if (k - (pow(t) - pow(t - 1)) * t > 0) {
      ans += (pow(t) - pow(t - 1));
      k -= (pow(t) - pow(t - 1)) * t;
    } else {
      ans += (k / t);
      k = 0;
    }
  }
  cout << ans;
}
