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;
long long tenll(long long n) {
  long long pp = 1;
  for (int i = 0; i < n; i++) pp *= 10;
  return pp;
}
long long w, m, k;
long long getdigit(long long num) {
  long long cnt = 0;
  while (num) {
    num /= 10;
    cnt++;
  }
  return cnt;
}
long long f(long long x) {
  long long dgm = getdigit(m);
  long long dgx = getdigit(x);
  dgx--;
  long long numm = tenll(dgm);
  long long numx = tenll(dgx);
  cout << "numx"
       << "->" << numx;
  cout << " ";
  cout << "numm"
       << "->" << numm;
  cout << endl;
  long long sum = (numm - m) * dgm * k + (x - numx + 1) * (dgx + 1) * k;
  cout << "sum"
       << "->" << sum;
  cout << endl;
  while (dgx >= dgm + 1) {
    sum += (tenll(dgx) - tenll(dgx - 1)) * dgx * k;
    dgx--;
  }
  cout << "sum"
       << "->" << sum;
  cout << endl;
  return sum;
}
int main() {
  cin >> w >> m >> k;
  long long lo = 0, hi = 1e9, mid;
  long long dg = getdigit(m);
  long long d = tenll(dg);
  long long cnt = 0;
  long long prev = m;
  while ((d - prev) <= w / (dg * k)) {
    cnt += (d - prev);
    w -= (d - prev) * dg * k;
    dg++;
    prev = d;
    d *= 10;
  }
  cnt += w / (dg * k);
  cout << cnt << endl;
}
