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<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef double dd;
const int N = 1e6+5;
const dd eps = 1e-8;
const int mod = 1e9+7;

ll a,b,n;
ll ans,ind;

int main(){
  ios::sync_with_stdio(false);
  // freopen("input.txt","r",stdin);
  cin>>a>>b>>n;
  for(ll i = n;i >= max(0ll,n-ll(N));i--){
    ll now = (a*i)/b - a*(i/b);
    if(now > ans){
      ans = now;
    }
  }
  cout<<ans<<endl;
  return 0;
}