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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define PI (acos(-1))
#define ct cout<<
#define cn cin>>
#define spc <<" "<<
#define nl puts("")
#define _fl(x) puts("FLAG " #x);
#define  _(x)  cout<< #x <<" is "<< x <<endl;
#define fs first
#define sc second
#define pb push_back
#define all(v)  (v).begin(), (v).end()
#define _109 (1000000000)
#define _0(arr) memset(arr,0,sizeof ( arr ) )
#define _1(arr) memset(arr,-1,sizeof ( arr ) )
#define _ninp(n,arr) for(int i=0; i<n;i++)cin>>arr[i];
#define _nout(n,arr) for(int i=0; i<n;i++)cout<<arr[i]<<" \n"[i==n-1];
double _eps=1e-6;
long long gcd(long long a, long long b) {
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

int main() {

    ll n,k;
    cn n>>k;

    if(n>=k){
     
       ll d=n/k;

       ll x=(n-d*k);
       ll p=abs(n-(d+1)*k);


       ct min(x,p);
    }
    else {

        ll x=abs(n-k);

    	ct min(x,abs(x-k));
    }



   nl;



    return 0;
}