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 <bits/stdc++.h>
using namespace std;
#define SPEED ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define fileio freopen("in.in", "r", stdin),freopen("out.out", "w", stdout);
#define ll long long int
#define FF first
#define SS second
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define pll pair<long long int,long long int>
#define sd(x) scanf("%d",&x)
#define slld(x) scanf("%lld",&x)
#define pd(x) printf("%d\n",x)
#define plld(x) printf("%lld\n",x)
#define pss printf
#define MOD 1000000007
#define INF 1e18
#define eps 0.00001
#define endl '\n'
#define debug(n1) cout<<n1<<endl
const int N = 1e5;  // limit for array size
ll n;  // array size

int main(){
	ll a,b,n;cin>>a>>b>>n;
	
	if(n>=b){
		cout<<floor((a*(b-1))/b)<<endl;
	}
	
	else{
		cout<<floor((a*(n))/b)<<endl;
	}
	
	return 0;
}