Problem:
Score : 400 points

Problem Statement

Given are integers A, B, and N.

Find the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N.

Here floor(t) denotes the greatest integer not greater than the real number t.

Constraints

1 ≤ A ≤ 10^{6}

1 ≤ B ≤ 10^{12}

1 ≤ N ≤ 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B N

Output

Print the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N, as an integer.

Sample Input 1

5 7 4

Sample Output 1

2

When x=3, floor(Ax/B)-A×floor(x/B) = floor(15/7) - 5×floor(3/7) = 2. This is the maximum value possible.

Sample Input 2

11 10 9

Sample Output 2

9

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < (n); i++)
#define REPR(i, n) for(int i = (n); i >= 0; i--)
#define FOR(i, m, n) for(int i = (m); i < (n); i++)
#define INF 1e9
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << #x << ": " << x << endl;

using namespace std;
typedef long long ll;

signed main() {
  ll a, b, n, ans;
  cin >> a >> b >> n;
  ans = floor((double)(a * n) / b) - a * floor((double)n / b);
  cout << ans << endl;
  return 0;
}