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


int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
//    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

//long long  factorial(long long n) {
//    if (n > 0) {
//        return n * factorial(n - 1);
//    } else {
//        return 1;
//    }
//}
long long  factorial(long long n, long long depth) {
    if (n > 1) {
        depth++;
        return factorial(n / 2, depth);
    } else {
        return depth;
    }
}

int min(int a, int b) {
    if (a > b)
        return b;
    else
        return a;
}

int distance(int a, int b) {
  return abs(a-b)*abs(a-b);
}



//int ()
/*
int triangle(int a, int b, int c){
  if(a<b+c){
    if(b<c+a){
      if(c<a+b){
        printf("1\n");
        return 1;
      }
    }
  }
  printf("1\n");
  return 0;
}
*/

int main()
{
//  int  a, b, c, d, e, X, Y ,r;
  unsigned long long  a, b, c, d, e, X, Y;
  //  double  c;
// long long  A[200003]={};
  int  A[200003]={};
  int  B[100003]={};
//  int  C[100003][100001]={};
  int  C[100003]={};

  int  flg=0;

//  char s[101][101]={};
  char s[100001]={};
  char t[100001]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
  long long p=0,q=0;
//  int flag=0;
//  int tmp=0;
//  char tmp[101]={};
//  long long min=10000000000;
//  long long min=100000001;
  int min =100000001;
  int len=0;
  int l=0;
//  int r=1000001;
  int cnt=0;
  int num=0;
  int num1=0;
  int num2=0;
  long long  ans=0;
  int ans_ac=0;
  int ans_wa=0;
  int max=0;
  long long sum=0;
  int addr=0;
  int ptn=0;
  int  tmp=0;
  double per=0; 
//  bool on_off[100]={};
  
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
  scanf("%lld %lld", &a, &b);
  a=a%b;
//  printf("%lld ",a);
  if((int)(a-b)<0) X=b-a;
  else X=a;

  
  
  printf("%lld\n", X);
  
return 0;
}