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 <math.h>
void A(void);
void B(void);
void C(void);
void C2(void);

int main(void){
    C();
    return 0;
}

void A(void){
    int a,b,c,k;
    scanf("%d %d %d",&a,&b,&c);
    printf("%d %d %d",c,a,b);
}

void B(void){
    int n,m,i,k=0;
    double sum=0;
    int array[1000];
    scanf("%d %d",&n,&m);
    for(i=0;i<n;i++){
        scanf("%d",&array[i]);
        sum+=array[i];
    }
    for(i=0;i<n;i++){
        if(array[i]>=sum/(4*m))k++;
    }
    if(k>=m){printf("Yes");
    }else{printf("No");}
}

void C(void){
    double a,b,def,min1,min2,min;
    int k=0;
    scanf("%lf %lf",&a,&b);
    min1=fmod(a,b);
    min2=fabs(fmod(a,b)-b);
    if(min1<min2){
        min=min1;
    }else{
        min=min2;
    }
    printf("%ld",(int)min);
}