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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP0(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define RREP0(i, n) for (int i = (n)-1; i >= 0; --i)
#define REP1(i, n) for (int i = 1, i##_len = (n); i <= i##_len; ++i)
#define RREP1(i, n) for (int i = (n); i >= 1; --i)
#define pow10(n) int(1e##n + n)

typedef long long LL;
typedef pair<int, int> P;

const int INTINF = int(1e9) + 1;
const LL LLINF = LL(1e18) + 1;
long double eps = 1.0E-14;

void solve()
{
    LL A, B, N;
    cin >> A >> B >> N;
    LL ans = 0;
    REP1 (x, min(N, A * 1000))
    {
        ans = max(ans, (A * x) / B - A * (x / B));
    }
    std::cout << ans << endl;
}

int main(int argc, char const *argv[])
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(20);
    solve();
    return 0;
}