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<stdio.h>
#define ll long long
ll llgcd(ll a, ll b){return b == 0 ? a : llgcd(b, a % b);}
ll lllcm(ll a, ll b){return a / llgcd(a, b) * b;}
int main(){
    long long a, b, n, ans = 0;
    scanf("%lld%lld%lld",&a,&b,&n);
    if(n < b)
        ans = n;
    else 
        ans = b - 1;
    ans *= a;
    ans /= b;
    printf("%lld\n",ans);
    return 0;
}