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>
using namespace std;
using ll = long long;

#define debug(x) (cerr << #x << " = " << x << endl;)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define FOR(it, b, e) for (typeof(b) it = (b); it != (e); ++it)
#define MSET(c, v) memset(c, v, sizeof(c))

const int INF = 0x3F3F3F3F; 
const int NEGINF = 0xC0C0C0C0;
const int NULO = -1;
const double EPS = 1e-10;

inline int cmp(double x, double y = 0) 
{  
  if(fabs(x - y) < EPS)
    return 0;
  return x > y ? 1 : -1;
}

ll a,b,n;

ll val(ll x)
{
  return floor((a * x ) / b) - a * floor(x / b);
}
  
int main()
{
  ios::sync_with_stdio(0);
  cin >> a >> b >> n;
  ll lo = 0;
  ll hi = n;
  while(hi - lo > 10)
  {
    ll mid1 = lo + (hi - lo) / 3;
    ll mid2 = hi - (hi - lo) / 3;
    if(val(mid1) < val(mid2))
    {
      lo = mid1;
    }
    else
    {
      hi = mid2;
    }
  }
  // cout << lo << ' ' << hi << endl;
  ll ans = 0;
  for(ll i = lo ; i <= hi; i++)
    ans = max(ans, val(i));
  cout << ans << endl;
  return 0;
}