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;
using namespace std;
long long sum(int i, long long* bit) {
  int s = 0;
  while (i > 0) {
    s += bit[i];
    i -= i & -i;
  }
  return s;
}
void add(int i, long long x, long long* bit, long long n) {
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}
int main() {
  double a, b;
  int n;
  while (~scanf("%lf %lf %d", &a, &b, &n)) {
    double c = a / b;
    long long* bit = new long long[n + 1]();
    c *= pow(10.0, n);
    int res = c;
    int idx = 1;
    while (res > 0) {
      add(idx++, res % 10, bit, n);
      res /= 10;
    }
    cout << sum(n, bit) << endl;
  }
}
