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;
unsigned typedef long long ll;
;
;
int mov[8][2] = {{0, -1}, {1, 0},  {-1, 0}, {0, 1},
                 {1, 1},  {1, -1}, {-1, 1}, {-1, -1}};
int knt[8][2] = {{1, 2}, {1, -2}, {-1, 2}, {-1, -2},
                 {2, 1}, {2, -1}, {-2, 1}, {-2, -1}};
ll m, k, w, ans;
int dij(ll x) {
  int res = 0;
  while (x > 0) {
    res++;
    x /= 10;
  }
  return res;
}
int main() {
  ll cost, nxt, cnt, tmp;
  cin >> w >> m >> k;
  ans = 0;
  cnt = dij(m);
  while (true) {
    nxt = 1;
    for (int i = 1; i <= cnt; i++) nxt *= 10;
    if (((nxt - m) * k * cnt) <= w) {
      w -= ((nxt - m) * k * cnt);
      ans += (nxt - m);
      m = nxt;
      cnt++;
    } else {
      ans += (w / (k * cnt));
      break;
    }
  }
  cout << ans << endl;
}
