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

typedef int keytype;

//quicksort(<#keytype *a#>, <#int first#>, <#int last#>)
//long long ;
void quicksort(keytype a[], int first, int last){
    int i, j;
    keytype x, t;
    
    x = a[(first + last) / 2];
    i = first;  j = last;
    for ( ; ; ) {
        while (a[i] < x) i++;
        while (x < a[j]) j--;
        if (i >= j) break;
        t = a[i];  a[i] = a[j];  a[j] = t;
        i++;  j--;
    }
    if (first  < i - 1) quicksort(a, first , i - 1);
    if (j + 1 < last) quicksort(a, j + 1, last);
}

int gcd(int x, int y){
    if (y == 0) return x;
    else        return gcd(y, x % y);
}

int main() {
    long long int a,b,m,n,k,ans,ans_2;
    scanf("%lld %lld",&n,&k);
    ans=n%k;
    ans_2=k-ans;
    if(n==0){
        printf("0\n");
    }else if (ans > ans_2) {
        printf("%lld\n",ans_2);
    }else{
        printf("%lld\n",ans);
    }
    return 0;
}