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 <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <iomanip>
#include <bitset>
#include <set>
#include <map>

#define rep(i,n) for (int i = 0; i < (n); i++)
#define FOR(i,a,b) for (int i=(a); i < (b); i++)
#define INF 100000000000
#define MOD 100000007 //10^9+7
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main()
{
    ll n, a, b, m, k, i, j;
    cin >> a >> b >> n;

    ll ans = 0LL;
    ll tmp;
    if ( a > b) {
        cout << n << endl;
        return 0;
    }
    for (i = 0; i <= n; i++) {
        tmp = floor((a * i)/b) - a*(floor(i/b));
        ans = max(ans, tmp);
    }

    cout << ans << endl;

    return 0;
}