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: Accepted)

#include<stdio.h>
#include<stdlib.h>
//#define scanf scanf_s
#define gc() getchar()
typedef unsigned long long ull;
typedef  long long ll;
int in() {   // 非負整数の入力
	int n = 0; int c;
	c = gc();
	do n = 10 * n + (c & 0xf); while ((c = gc()) >= '0');
	return n;
}
int comp(const void* a, const void* b)//用来做比较的函数。
{
	return *(int*)a - *(int*)b;
}
ll qabs(ll a, ll b)
{
	if (a - b < 0) return b - a;
	else return a - b;
}
int main()
{
	ll n, k, m, t;
	scanf("%lld %lld", &n, &k);
	if (n == k) { printf("0\n"); return 0; }
	if (n < k) n += k;
	m = n % k;
	t = qabs(m , k);
	printf("%lld", m < t ? m : t);
}