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: Runtime Error)

#include <stdio.h>
#include <math.h>

int main() {
    long A, B, N;
    scanf("%ld %ld %ld", &A, &B, &N);
    int x[N+1], m = 0;

    for (int i = 0; i <= N; i++) {
        x[i] = floor(A * i / B) - A * floor(i / B);
        //printf("%d\n", x[i]);
    }
    for (int i = 0; i <= N; i++) {
        if (m < x[i]) {
            m = x[i];
        }
    }
    
    printf("%d\n", m);
    return 0;
}