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 <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define Rep(i,a,n)for(int i=(int)(a);i<(int)(n);i++)
#define rep(i,n)Rep(i,0,n)
typedef long long ll;
typedef long double ld;
#define M(x,y) ((x)>(y)?(x):(y))
#define m(x,y) ((x)<(y)?(x):(y))

int main() {
	ll a,b,n,ans=0;
	scanf("%lld %lld %lld", &a,&b,&n);
	if (b % a == 0) {
		if (n < b)ans = M(ans, (a * n / b));
		else ans = M(ans, (a * (b - 1) / b));
	}
	Rep(i,1,a+1) {
		ll x=floorl(i*b/a);
		if (x > n) break;
		ans = M(ans,floorl(a*(x%b)/b));
	}
	ans = M(ans, floorl(a*(n % b)/b));
	ll Ans = ans;
	printf("%lld", Ans);
}