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: Runtime Error)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <unordered_set>
#pragma GCC optimize("Ofast")
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int, int>> vpii;
typedef vector<vector<int>> vvi;
typedef vector<vector<char>> vvc;
typedef vector<vector<string>> vvs;
typedef vector<vector<ll>> vvll;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define irep(it, stl) for (auto it = stl.begin(); it != stl.end(); it++)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define fin(ans) cout << (ans) << '\n'
#define mp(p, q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Sort(a) sort(a.begin(), a.end())
#define Rort(a) sort(a.rbegin(), a.rend())
#define MATHPI acos(-1)
#define itn int;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
struct io
{
    io()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
    }
};
const int INF = INT_MAX;
const ll LLINF = 1LL << 60;
const ll MOD = 1000000007;
const double EPS = 1e-9;

#include <algorithm>
#include <functional>
#include <iostream>

#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))


int main(void)
{
    int A, B, N;
    int Z = 0;
    int Y = 0;
    cin >> A;
    cin >> B;
    cin >> N;
    vi retsu(N, 0);

    for (int x = 1; x <= N; x++)
    {
        
        
        Z = A * x / B;
        Y = x / B;
        //printf("%d,%d",Z,Y);
        retsu[x]=Z-Y;
        //printf("%d\n",retsu[x]);
    }
    
    std::sort(std::begin(retsu),std::end(retsu),greater<int>());
    printf("%d",retsu[0]);
}