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: Accepted)

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

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define MOD 1000000000 + 7
#define PI 3.14159265359

const long long INF = 100000;
typedef pair<int,int> P;
typedef long long ll;

using Graph = vector<vector<int>>;
ll A, B;
ll func(ll x) {
	return A*x / B - A * (x/B);
}
int main() {
	ll N;
	cin >> A >> B >> N;
	cout << func(min(N, B - 1)) << endl;
}