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
// clang-format off
#include <bits/stdc++.h>
#define int long long
#define main signed main()
// #define main int main()
#define loop(i, a, n) for (int i = (a); i < (n); i++)
#define rep(i, n) loop(i, 0, n)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define prec(n) fixed << setprecision(n)
#define stlice(from, to) substr(from, (to) - (from) + 1)
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
using namespace std;
using pii = pair<int, int>;
using vi = vector<int>;
using vd = vector<double>;
using vc = vector<char>;
using vb = vector<bool>;
using vs = vector<string>;
using vpii = vector<pii>;
using vvi = vector<vi>;
using vvb = vector<vb>;
using vvpii = vector<vpii>;
template<typename A> using fn = function<A>;
constexpr int INF = sizeof(int) == sizeof(long long) ? 1000000000000000000LL : 1000000000;
constexpr int MOD = 1000000007;
constexpr double PI = acos(-1);
template<typename A, typename B> bool cmin(A &a, const B &b) { return a > b ? (a = b, true) : false; }
template<typename A, typename B> bool cmax(A &a, const B &b) { return a < b ? (a = b, true) : false; }
constexpr bool odd(const int &n) { return n & 1; }
constexpr bool even(const int &n) { return !odd(n); }
void solve();
main { solve(); return 0; }
// clang-format on

void solve() {
  int a, b, n;
  while (cin >> a >> b >> n) {
    double x = double(a) / double(b);
    auto f   = [&](int i) { return ((int) (x * pow(10, i))) % 10; };
    int s    = 0;
    loop(i, 1, n + 1) s += f(i);
    cout << s << endl;
  }
}