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 T GCD(T a, T b) {
  return b ? GCD(b, a % b) : a;
}
template <class T>
inline T LCM(T a, T b) {
  if (a < 0) return LCM(-a, b);
  if (b < 0) return LCM(a, -b);
  return a * (b / GCD(a, b));
}
template <class T>
inline T POW1(T a, T b, T m) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > m) x %= m;
    }
    y = (y * y);
    if (y > m) y %= m;
    b /= 2;
  }
  return x;
}
template <class T>
inline T INV(T n, T m) {
  return POW1(n, m - 2, m);
}
template <class T>
inline T SUB(T a, T b, T m) {
  return (a % m - b % m + m) % m;
}
template <class T>
inline T ADD(T a, T b, T m) {
  return (a % m + b % m) % m;
}
template <class T>
inline T MUL(T a, T b, T m) {
  return (a % m * b % m) % m;
}
template <class T>
inline T DIV(T a, T b, T m) {
  return (a % m * (INV(b, m)) % m) % m;
}
int main() {
  long long int w, m, k;
  cin >> w >> m >> k;
  long long int tm = m;
  long long int d1 = 0;
  while (tm != 0) {
    tm /= 10;
    d1++;
  }
  long long int sum = 0, prev = 0;
  long long int ans = 0;
  long long int nxt = 1, temp = d1;
  for (long long int i = 0; i < d1; i++) {
    nxt *= 10;
  }
  long long int num = m, prev_n;
  while (sum <= w) {
    prev = sum;
    prev_n = nxt - num;
    ans += prev_n;
    if ((log(prev_n) + log(k) + log(d1)) / log(10) > 16) {
      d1++;
      break;
    }
    sum += prev_n * d1 * k;
    d1++;
    num = nxt;
    nxt *= 10;
  }
  d1--;
  ans -= prev_n;
  ans += (w - prev) / (d1 * k);
  cout << ans << endl;
  return 0;
}
