#include <bits/stdc++.h>
using namespace std;
long long w, m, k;
int s(long long n) {
  char buf[200];
  sprintf(buf, "%I64d", n);
  return strlen(buf);
}
long long ans = 0;
int main() {
  scanf("%I64d %I64d %I64d", &w, &m, &k);
  long long sm = s(m);
  long long ws = pow(10, sm);
  w /= k;
  long long t = ws - m;
  if (t * sm <= w) {
    ans += t;
    w -= t * sm;
    long long wsn = ws * 10;
    sm += 1;
    t = wsn - ws;
    while (t * sm <= w) {
      ans += t;
      w -= t * sm;
      ws = wsn;
      wsn = ws * 10;
      sm += 1;
      t = wsn - ws;
    }
    ans += w / (sm);
  } else {
    ans = w / (sm);
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long w, m, k;
int s(long long n) {
  char buf[200];
  sprintf(buf, "%I64d", n);
  return strlen(buf);
}
long long ans = 0;
int main() {
  scanf("%I64d %I64d %I64d", &w, &m, &k);
  long long sm = s(m);
  long long ws = pow(10, sm);
  w /= k;
  long long t = ws - m;
  if (t * sm <= w) {
    ans += t;
    w -= t * sm;
    long long wsn = ws * 10;
    sm += 1;
    t = wsn - ws;
    while (t * sm <= w) {
      ans += t;
      w -= t * sm;
      ws = wsn;
      wsn = ws * 10;
      sm += 1;
      t = wsn - ws;
    }
    ans += w / (sm);
  } else {
    ans = w / (sm);
  }
  printf("%I64d\n", ans);
  return 0;
}

```