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)

#include"bits/stdc++.h"
using namespace std;
#define REP(k,m,n) for(int (k)=(m);(k)<(n);(k)++)
#define rep(i,n) REP((i),0,(n))
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using tp3 = tuple<int, int, int>;
using Mat = vector<vector<ll>>;
constexpr int INF = 1 << 28;
constexpr ll INFL = 1ll << 60;
constexpr int dh[4] = { 0,1,0,-1 };
constexpr int dw[4] = { -1,0,1,0 };
bool isin(const int H, const int W, const int h, const int w) {
    return 0 <= h && h < H && 0 <= w && w < W;
}
template<typename T>
T minch(T& l, T r) {
    return l = min(l, r);
}
template<typename T>
T maxch(T& l, T r) {
    return l = max(l, r);
}
template<typename T>
void output(const T& val) {
    cout << val << endl;
}
template<typename T>
void output(const vector<T>& vec, const bool newline = false) {
    for (const T& val : vec)cout << val << (newline ? '\n' : ' '); cout << endl;
}
template<typename T>
void output(const vector<vector<T>>& mat) {
    for (const auto& row : mat)output(row);
}
// ============ template finished ============

int main()
{
    ll A, B, N;
    cin >> A >> B >> N;
    ll times = N / B;
    if (times == 0) {
        ll l = A * N / B;
        ll r = A * (N / B);
        cout << l - r << endl;
    }
    else {
        ll x = B * times - 1;
        ll l = A * x / B;
        ll r = A * (x / B);
        cout << l - r << endl;
    }
    return 0;
}