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 int long long
#define mod 1000000009
#define garou ios_base::sync_with_stdio(false); cin.tie(NULL);
using namespace std;
int pow(int p,int r){
	int ans=1;
	while(r>0){
     if(r&1){
		 ans=((ans%mod)*(p%mod))%mod;
	 }
	 r=r>>1;
     p=((p%mod)*(p%mod))%mod;
	}
}
signed main() {
	garou;
    int a,b,n;
	cin>>a>>b>>n;
	int l=0,r=n,mx=INT_MIN;
    while(r>=l){
		int mid=(l+r)/2;
		int o=floor((a*mid)/b)-a*floor(mid/b);
		if(o<=mx){
			r=mid-1;
		}else if(o>mx){
			mx=max(mx,o);
			l=mid+1;
		}
	}
	cout<<mx;
}