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 GCC optimize("O3")
using namespace std;
long long w, m, k;
long long s(long long x) {
  long long cnt = 0;
  while (x > 0) {
    x /= 10;
    cnt++;
  }
  return cnt;
}
long long limit(long long st, long long en, long long c) {
  if (c / st + 2 < en) return 1;
  return 0;
}
bool check(long long x, long long w, long long m, long long k) {
  if (s(x) == s(m)) {
    if (limit(x - m + 1 * s(x), k, w)) return 0;
    return (((x - m + 1) * s(x) * k) <= w);
  }
  long long sum = 0, p = 9;
  for (int i = 1; i <= s(m); i++) {
    p *= 10;
  }
  for (int i = s(m) + 1; i <= s(x) - 1; i++) {
    sum += p * i * k;
    if (sum > w) return 0;
    p *= 10;
  }
  p = 1;
  for (int i = 1; i <= s(m); i++) {
    p *= 10;
  }
  sum += (p - m) * s(m) * k;
  p = 1;
  for (int i = 1; i < s(x); i++) {
    p *= 10;
  }
  sum += (x - p + 1) * s(x) * k;
  if (sum < 0) {
    return 0;
  }
  return sum <= w;
}
int main() {
  cin >> w >> m >> k;
  long long tl = m, tr = 1e17;
  while (tr - tl > 1) {
    long long mid = (tr + tl) >> 1;
    if (check(mid, w, m, k))
      tl = mid;
    else
      tr = mid;
  }
  if (!check(tl, w, m, k))
    cout << 0;
  else {
    if (check(tr, w, m, k)) tl = tr;
    cout << (tl - m + 1) << '\n';
  }
  return 0;
}
