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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long pow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x);
    x = (x * x);
    n >>= 1;
  }
  return res;
}
long long w, m, k;
bool predicate(long long mid) {
  long long x1 = m, x2 = m + mid;
  unsigned long long sum = 0;
  vector<long long> A;
  A.push_back(x1);
  A.push_back(x2);
  string s1, s2;
  long long cnt = 0;
  s1 = to_string(x1);
  s2 = to_string(x2);
  long long x = s1.size();
  long long y = s2.size();
  while (y > x) {
    A.push_back(pow(10, y - 1));
    y--;
  }
  sort(A.begin(), A.end());
  for (long long i = 0; i < A.size() - 1; i++) {
    sum += abs(A[i] - A[i + 1]) * k * x;
    x++;
  }
  if (sum <= w) return true;
  return false;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> w >> m >> k;
  long long l = 0, h = w;
  long long mid = 0, ans = INT_MIN;
  while (l <= h) {
    mid = l + (h - l) / 2;
    if (predicate(mid)) {
      l = mid + 1;
      ans = max(ans, mid);
    } else
      h = mid - 1;
  }
  cout << ans << '\n';
  return 0;
}
