Problem:
Score : 300 points

Problem Statement

Given any integer x, Aoki can do the operation below.

Operation: Replace x with the absolute difference of x and K.

You are given the initial value of an integer N. Find the minimum possible value taken by N after Aoki does the operation zero or more times.

Constraints

0 ≤ N ≤ 10^{18}

1 ≤ K ≤ 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the minimum possible value taken by N after Aoki does the operation zero or more times.

Sample Input 1

7 4

Sample Output 1

1

Initially, N=7.

After one operation, N becomes |7-4| = 3.

After two operations, N becomes |3-4| = 1, which is the minimum value taken by N.

Sample Input 2

2 6

Sample Output 2

2

N=2 after zero operations is the minimum.

Sample Input 3

1000000000000000000 1

Sample Output 3

0

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

typedef long long int llint;

llint mod_calc(llint n, llint k)
{
	llint val = n;
	llint mod = 0;

	while(1) {
		mod = val % k;
		printf("mod %d\n", mod);
		if (mod != 0) break;
		if (val <= k) break;
		val = val / k;
	}

	return mod;
}

llint abs_calc_mul(llint val1, llint val2)
{
	return (val1 > val2)? (val1 - val2) : (val2 - val1);
}

int main()
{
	llint n = 0, k = 0;
	llint val = 0, ret = 0;
	if (scanf("%lld %lld", &n, &k));
	if (n == k) goto func;
	if (k == 1) goto func;

	if (n < k) {
		if (n == 0) {
			val = k;
			goto func;
		}
		val = abs_calc_mul(n, k);
		ret = abs_calc_mul(val, k);
		val = (val < ret) ? val : ret;
	} else {
		val = n % k;
		ret = abs_calc_mul(val, k);
		val = (val < ret) ? val : ret;		
	} 

func:
	printf("%lld", val);

	return 0;

}