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)

/*	
*  Sachin Mittal
*/
#include<bits/stdc++.h>
#define int long long
#define M 998244353
#define mod 1000000007
#define all(v) v.begin(),v.end()
#define rep(i,j,n) for(int i=j;i<n;i++)
#define F first
#define S second
#define lb lower_bound
#define mem0(dp) memset(dp,0,sizeof(dp))
#define mem1(dp) memset(dp,-1,sizeof(dp))
#define pb push_back
#define sz(s) (int)(s.size())
#define debug(x) cout << #x << " = " << x << '\n';
using namespace std;

void solve()
{
   	int a, b, n;
   	cin >> a >> b >> n;
   	int prod = a * n;
   	prod = prod - prod % b;
   	int ans = prod / a ;
   	ans = n;
   	int ans1=(ans * a)/b - a*(ans/b);
   	if(ans+1<=n){
   		ans++;
   		ans1=max(ans1,(ans * a)/b - a*(ans/b));
   	}
   	cout<<ans1;

}

int32_t main(){
ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int tst=1;
//cin>>tst;
while(tst--)
{
	solve();
}
}