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;

#define rep(i, n) for (int i = 0; i < n; ++i)
#define all(x) (x).begin(),(x).end()

using ll = long long;
using P = pair<int,int>;

const int MOD = 1e9 + 7;
const int INF = 1001001001;


int main()
{
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

    ll a, b, n;
    cin >> a >> b >> n;

    ll ans = max((a * n) / b - a * (n / b), (a * (n - 1)) / b - a * ((n - 1) / b));
    ans = max(ans, (ll)0);

    cout << ans << "\n";

    return 0;
}