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)


#define _USE_MATH_DEFINES

#include <vector>
#include <list>
#include <map>
#include <string>
#include <iostream>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <utility>
#include <ctgmath>
#include <queue>
#include <set>

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;

const ll inf = 10e17;
const ll mod = inf;//1000000007;
const ld eps = 10e-9;



ll solve(void);
ll gcd(ll x, ll y);


ll mulmod(ll x, ll y)
{
	return (x * y) % mod;
}

ll submod(ll x, ll y)
{
	return (x - y + 100 * mod) % mod;
}

ll power(ll n, ll r)
{
	if ( r == 0 ) { return 1; }
	if ( r == 1 ) { return n; }
	if ( n == 0 ) { return 0; }
	if ( n == 1 ) { return 1; }

	if (r % 2 == 0)
	{
		ll y = power(n, r/2);
		return mulmod(y, y);
	}
	else
	{
		ll y = power(n, (r-1)/2);
		return n * mulmod(y, y) % mod;
	}
}

ll gcd(ll x, ll y)
{
	if (y == 1) { return 1;}
	if (x % y == 0) { return y; }
	return gcd(y, x % y);
}

vector<ll> perm;
vector<ll> inv;
void init_inv(ll m)
{
	perm.resize(m);
	inv.resize(m);
	
	perm[0] = 1;
	inv[0] = 1;

	for (ll i = 1; i < m; i++)
	{
		perm[i] = mulmod(perm[i-1], i);
	}
	
	for (ll i = 1; i < m; i++)
	{
		inv[i] = power(perm[i], mod - 2);
	}
}

ll ncr(ll n, ll r)
{
	// n (n-1) ... (n-r)
	if ( r == n || r == 0 ) { return 1; }

	ll ans = 1;
	for (ll i = 0; i < r; i++)
		ans = mulmod(ans, n - i);

	 return mulmod(ans, inv[r]);
	// return mulmod(perm[n], mulmod(inv[n-r], inv[r]));
}

bool binary_check(ll mid, ll key)
{
	return mid >= key;
}

ll binary_search()
{
//	ll ng = 0;
//	ll ok = x;
//
//	while (abs(ok - ng) > 1)
//	{
	//	ll mid = (ok+ng) / 2;
	//	if (binary_check(mid, k)) ok = mid;
	//	else ng = mid;
	//}

	return 0;
	//return ok;
}

ll a, b, n;

ll solve(void)
{
	cin >> a >> b >> n;

	ll k = (n * a) / b;

	cout << min(a, k) << endl;

	return 0;
}

int main(void)
{
	while (solve())
	{
	}
	return 0;
}