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<stdio.h>
#include<string>
#include<vector>
#include<map>
#include<tuple>
#include<algorithm>
#include<cmath>
#include<limits>
#include<set>
#include<deque>
#include<queue>
using namespace std;
#define int long  long int
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
typedef pair<int, int>P;
const int MOD = 1e9 + 7;
//const int MOD = 998244353;
const int INF = 1e11;
const long double PI = (acos(-1));

int A, B, N;

int f(int n) {
	int a = A * n / B;
	int b = n / B;
	int c = A * b;
	return a - c;
}

signed main() {
	cin >> A >> B >> N;
	int num;
	if (N % B == 0)num = N - 1;
	else num = N;
	if (N == 1)num = 1;
	cout << f(num);
}