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;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
#define Rep(i,n) for(int i=1;i<=n;i++)
const ll MOD = 1000000007;
long long modinv(long long a, long long m=MOD) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m; 
    if (u < 0) u += m;
    return u;
    /*MODの割り算の際に
    a/＝b;
    を
    a *= modinv(b,MOD);
    とする。
        */
}
int main(){
    int A,B,N;cin >> A >> B >> N;

    ll min = 0;
    ll max = N+1;
    ll ans = 0;
    ll mid;
    while(1){
        mid = (min+max)/2;
        ll tmp = (ll)((A* mid)/B) - A*(ll)(mid/B);
        if(tmp>ans){
            ans = tmp;
            min = mid;
        }else{
            max = mid;
        }
        if(abs(max-min)<=1)break;
    }

    cout << ans << endl;

    return 0;
}