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<algorithm>
#include<map>
#include<set>
#include<queue>
#include<iomanip>
#include<sstream>
#include<cmath>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define req(i,n) for(int i = 1;i <=  n; i++)
#define rrep(i,n) for(int i = n -1;i >= 0;i--)
typedef long long int ll;
typedef long double ld;
const int inf = (1 << 30);
const int MOD = 1000000007;
template<typename A, size_t N, typename T>
void Fill(A(&array)[N], const T & val) {
    std::fill((T*)array, (T*)(array + N), val);
}

int main() {
    ll a, b, n; cin >> a >> b >> n;
    if (n < b) {
        cout << (a * n) / b - a * (n / b) << endl;
    }
    else {
        cout << a - 1 << endl;
    }
}