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 <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <deque>

//using namespace std;
using std::cin, std::cout, std::endl, std::string;
using ll = long long;

#define rep(countName, left, right, up) for (auto countName = left; countName < right; countName += up)
#define REP(countName, end) for (auto countName = 0; countName < end; countName++)
#define rev(countName, right, left, down) for (auto countName = right; countName >= left; countName -= down)
#define REV(countName, end) for (auto countName = end; countName >= 0; countName--)

auto solve1(ll a, ll b, ll n)
{
    ll ans = 0;
    ans = std::max(ans, ((a * (n - b - 1)) / b) - a * ((n - b - 1) / b));
    ans = std::max(ans, ((a * n) / b) - a * (n / b));
    return ans;
}

int main(void)
{
    ll a, b, n;
    cin >> a >> b >> n;
    cout << solve1(a, b, n) << endl;

    return 0;
}