Assume that a, b, and n are all positive integers. Let f (i) be the i-th fraction of the fraction a / b (0 ≤ f (i) ≤ 9). At this time, let s be the sum of f (i) from i = 1 to n.

s = f (1) + f (2) + ... + f (n)


Create a program that reads a, b, n, outputs s, and exits.



Input

The input consists of multiple datasets. For each dataset, three integers a (1 ≤ a ≤ 1000), b (1 ≤ b ≤ 10000), n (1 ≤ n ≤ 100) are given on one line, separated by blanks.

The number of datasets does not exceed 100.

Output

Prints s on one line for each dataset.

Example

Input

1 2 3
2 3 4
5 4 3
4 3 2


Output

5
24
7
6
#include <bits/stdc++.h>
using namespace std;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
void preview(T a, T b) {
  for (T it = a; it != b; ++it) cerr << *it << " ";
  cerr << endl;
}
const int INF = 100000000;
const double PI = acos(-1.0), EPS = 1e-10;
int main() {
  for (int a, b, n; cin >> a >> b >> n;) {
    double f = (double)a / b;
    char buf[1024];
    sprintf(buf, "%.30f", f);
    string s(buf);
    int start = find((s).begin(), (s).end(), '.') - s.begin() + 1;
    vector<int> sum;
    for (int i = 0; i < (int)n; ++i) {
      sum.push_back(s[start + i] - '0');
    }
    cout << accumulate((sum).begin(), (sum).end(), 0) << endl;
  }
  return 0;
}
