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;

int main()
{


        long long n,m,a,b,c,d,e,f,g,h,p,q,r,fr,sc,tr,sz,tz,sz1,sz2,i,j,k,mx=LLONG_MIN,mn=LLONG_MAX; string str,ttr,ntr;
        long long x=0,y=0,cnt=0,res=0,ttl=0,ans=0,sum=0; long long flg=0,flag=1,na=0,as=1;
        vector<long long>u,v,w; vector< pair<long long,long long> >vct,tct; vector<string>vst;
        set<long long>st,nt,tt; map<long long,long long>mp,nq,qr,mp1,mp2;

        scanf("%lld %lld %lld",&a,&b,&n);
        fr=min(b-1,n);
        sc=a*fr;  p=sc/b;  printf("%lld\n",p);







    return 0;
}