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;

const double PI = 3.14159265358979323846;
typedef long long ll;
const double EPS = 1e-9;
#define rep(i, n) for (int i = 0; i < (n); ++i)
//#define rep(i, n) for (ll i = 0; i < (n); ++i)
const ll INF = 10e17;
#define cmin(x, y) x = min(x, y)
#define cmax(x, y) x = max(x, y)
#define ret() return 0;

double equal(double a, double b) {
    return fabs(a - b) < DBL_EPSILON;
}

std::istream &operator>>(std::istream &in, set<string> &o) {
    string a;
    in >> a;
    o.insert(a);
    return in;
}

std::istream &operator>>(std::istream &in, queue<int> &o) {
    ll a;
    in >> a;
    o.push(a);
    return in;
}

typedef pair<ll, ll> P;

bool contain(set<int> &s, int a) { return s.find(a) != s.end(); }

//ofstream outfile("log.txt");
//outfile << setw(6) << setfill('0') << prefecture << setw(6) << setfill('0') << rank << endl;
// std::cout << std::bitset<8>(9);
//const ll mod = 1e10;

typedef priority_queue<ll, vector<ll>, greater<ll> > PQ_ASK;


int main() {

    ll a, b, n;
    cin >> a >> b >> n;
    ll ma = max(1ll, ll(n - 10e7));
    ll ans = -INF;
    for (ll x = n; x >= ma; x--) {
        ll now = (a * x / b) - a * (x / b);
        cmax(ans, now);
    }

    ll ma2 = min(n, 2 * a);
    for (ll x = 1; x < ma2; x++) {
        ll now = (a * x / b) - a * (x / b);
        cmax(ans, now);
    }

    cout << ans << endl;


}