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 rep(i,n) for(int i = 0; i < n; i++)
#define vi vector<int>
#define vvi vector<vi>
#define pii pair<int, int>
#define mod 1000000007
using ll = long long;
using namespace std;
int main(){
  ll a,b,n;
  cin >> a >> b >> n;
  ll ans = 0;
  for(ll x = n; x >= 0; x--){
    ll val = ll(a * x / b) - a * ll(x / b);
    ans = ans < val ? val : ans;
    if(ans > val + 10) break;
  }
  cout << ans << endl;
}