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++ (Clang 10.0.0) (judge verdict: Accepted)

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cmath>
#include<queue>
#include<vector>
#include<sstream>
#include<algorithm>
#include<stack>
#include<bitset>
#include<map>
#include<set>
#define rep(i,n) for(int i=0;(i)<(n);i++)
#define rep1(i,n) for(int i=1;(i)<=(n);i++)
#define mst(a,b) memset(a,b,sizeof(a))
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define dbg(a) cout<<"* "<< #a <<" : "<<a<<endl
#define fr first
#define se second
#define ls x<<1
#define rs x<<1|1
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define lowbit(x) x&(-x)
#define ac cout<<ans<<endl
//#define DEBUG 0
using namespace std;
const double pi=acos(-1.0);
const double eps=1e-8;
const ull hashp=131;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int inf=0x3f3f3f3f;
const int mod=998244353;
const int maxn=2e5+100;

void solve(){
	ll a,b,n;
	cin>>a>>b>>n;
	ll nxt=n-n%b-1;
	if(nxt<0)nxt=0;
	printf("%lld",max(a*n/b-a*(n/b),a*nxt/b-a*(nxt/b)));
}
int main(){
    //IOS;
    //freopen("in.txt","r",stdin);
    //freopen("D:\\data.txt","w",stdout);
    solve();
    
	;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
	;;                                             ;;
	;;         ;;            ;;;;;;;;;       ;;    ;;
	;;        ;; ;;        ;;                ;;    ;;
	;;       ;;   ;;       ;;                ;;    ;;
	;;      ;;;;;;;;;      ;;                ;;    ;;
	;;     ;;       ;;     ;;                      ;;
    ;;    ;;         ;;      ;;;;;;;;;       ;;    ;;
    ;;                                             ;;
    ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
    
    return 0;
}