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 int long long
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rep1(i,n) for(int i=1;i<=(int)(n);++i)
#define rep11(i,n) for(int i=1;i<(int)(n);++i)
#define repo(i,o,n) for(int i=o;i<(int)(n);++i)
#define repm(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define pb(n) push_back(n)
#define mp make_pair
#define INF LONG_LONG_MAX
#define MOD 1000000007
#define yn(f) cout << (f ? "Yes" : "No") << endl;
#define YN(f) cout << (f ? "YES" : "NO") << endl;

int a,b,n,ans;


signed main() {
    clock_t start = clock();
    cin >> a >> b >> n;
    rep(i,min(n,2000000LL)) ans = max(ans,(a*i)/b - a*(i/b));
    int i=b-1;
    while(clock() - start <= 1.8*CLOCKS_PER_SEC) {
        if (i > n) break;
        ans = max(ans,(a*i)/b - a*(i/b));
        i += b;
    }
    cout << ans << endl;
}