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 vll vector<ll>
#define vi vector<int>
#define vpll vector<pair<ll,ll>>
#define pii pair<int,int>
#define REP(i,a) for(int (i)=0; (i)<(a); (i)++)
#define RREP(i,a) for(int (i)=a-1; (i)>=(0); (i)--)
#define REP2(i,a,b) for(int (i)=(a); (i)<(b); (i)++)
#define MOD 1000000007
#define ALL(v) v.begin(), v.end()
#define m0(x) memset(x,0,sizeof(x))
typedef long long ll;
using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll A,B,N; cin >> A >> B >> N;
    ll ans = (ll)(A*N/B) - A*(ll)(N/B);
    REP(i,A){
    N--;
    if(N==0)break;
    ll tmp = (ll)(A*N/B) - A*(ll)(N/B);
    ans = max(tmp, ans);
    }
  	return 0;
}