#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
int count(long long int n) {
  int c = 0;
  while (n) {
    n /= 10;
    c++;
  }
  return c;
}
int main() {
  long long int w, m, k;
  cin >> w >> m >> k;
  int cur = count(m);
  long long int p = 1;
  long long int tmp = cur;
  while (count(p) <= cur) p *= 10;
  long long int ans = 0;
  while (w) {
    long long int s = p - m;
    long long int add = min(w / (k * cur), s);
    w -= add * k * cur;
    if (add == 0) break;
    ans += add;
    cur++;
    m = p;
    p *= 10;
  }
  cout << ans;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize("Ofast")
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
int count(long long int n) {
  int c = 0;
  while (n) {
    n /= 10;
    c++;
  }
  return c;
}
int main() {
  long long int w, m, k;
  cin >> w >> m >> k;
  int cur = count(m);
  long long int p = 1;
  long long int tmp = cur;
  while (count(p) <= cur) p *= 10;
  long long int ans = 0;
  while (w) {
    long long int s = p - m;
    long long int add = min(w / (k * cur), s);
    w -= add * k * cur;
    if (add == 0) break;
    ans += add;
    cur++;
    m = p;
    p *= 10;
  }
  cout << ans;
}

```