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)

/*Read the problem carefully before starting to work on it*/
#include <bits/stdc++.h>
//#include <boost/multiprecision/cpp_int.hpp>
//using namespace boost::multiprecision;
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;

typedef long long ll;

#define pb push_back
#define eb emplace_back
#define mp(x,y) make_pair(x,y)
#define mod 1000000007

double PI = 3.1415926535897932384626;

//template<typename T> T power(T x,T y,ll m=mod){T ans=1;while(y>0){if(y&1LL) ans=(ans*x)%m;y>>=1LL;x=(x*x)%m;}return ans%m;}

//typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> ost;

//mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
//#define rnd(x, y) uniform_int_distribution<ll>(x, y)(rng)

//struct custom_hash {
//    static uint64_t splitmix64(uint64_t x) {
//        // http://xorshift.di.unimi.it/splitmix64.c
//        x += 0x9e3779b97f4a7c15;
//        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
//        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
//        return x ^ (x >> 31);
//    }
//    size_t operator()(uint64_t x) const {
//        static const uint64_t FIXED_RANDOM =
//            chrono::steady_clock::now().time_since_epoch().count();
//        return splitmix64(x + FIXED_RANDOM);
//    }
//};

#define fi first
#define se second

int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll a, b, n;
    cin >> a >> b >> n;
    ll hehe = min(n, b - 1);
    ll one = (a * hehe) / b;
    ll two = a * (hehe / b);
    cout << (one - two);
    return 0;
}


//252908XL

/*
comments:-

*/