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 <iostream>
#include <iomanip>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <numeric>
#include <random>
#include <algorithm>
#include <functional>
#include <cassert>

using namespace std;
typedef long long ll;

template<class T> inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template<class T> inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

const int INF = (1 << 30) - 1;
const ll INFLL= (1LL << 61) - 1;
const int MOD = 1000000007;
#define ALL(a) (a).begin(),(a).end()
#define rALL(a) (a).rbegin(),(a).rend()
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

int main(){
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  ll A,B,N;
  cin>>A>>B>>N;
  ll ans=0;
  REP(i,2*A){
    if(i>N)break;
    chmax(ans,A*i/B - A*(i/B));
  }
  cout<<ans<<endl;
}