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 vec8[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1},
                  {0, 1},   {1, -1}, {1, 0},  {1, 1}};
int vec4[4][2] = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
string int_to_string(int n, int base = 10) {
  string s = "";
  do {
    s += char((n % base) + '0');
    n /= base;
  } while (n > 0);
  reverse(s.begin(), s.end());
  return s;
}
int gcd(int uno, int dos) {
  if (dos == 0) return uno;
  return gcd(dos, uno % dos);
}
int mcm(int first, int second) { return (first / gcd(first, second)) * second; }
int Orientacion(pair<int, int> p, pair<int, int> q, pair<int, int> r) {
  return (q.first * r.second + r.first * p.second + p.first * q.second) -
         (q.first * p.second + r.first * q.second + p.first * r.second);
}
long long int w, m, k, pot, dig, pagar, resp, nueve;
int main() {
  int i, j, n, casos, cont;
  while (cin >> w >> m >> k) {
    pot = 10;
    resp = 0;
    for (dig = 1, nueve = 9; w > 0; dig++, pot *= 10) {
      if ((pot - 1) >= m) {
        pagar = w / (dig * k);
        if (pagar == 0) break;
        if (pagar <= ((pot - 1) - m + 1)) {
          resp += pagar;
          break;
        } else {
          w -= (((pot - 1) - m + 1) * dig * k);
          resp += ((pot - 1) - m + 1);
        }
        m = pot;
      }
      nueve *= 10;
    }
    cout << resp << endl;
  }
  return 0;
}
