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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < n; i++)
#define REP1(i,n) for(int i = 1; i < n+1 ;i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define INF 2e9
#define ALL(v) v.begin(), v.end()
#define SZ(x) ((int)(x).size())
#define ARRAY_LENGTH(array) (sizeof(array) / sizeof(array[0]))
using namespace std;
typedef long long ll;
using p = pair<int, int>;
using P = pair<ll, ll>;
using V = vector<int>;
using VV = vector<vector<int>>;
using VVV = vector<vector<vector<int>>>;
using VL = vector<ll>;
using VVL = vector<vector<ll>>;
using VVVL = vector<vector<vector<ll>>>;
using Graph = VV;

int main(){
    ios::sync_with_stdio(false); cin.tie(0);
    ll n,k;
    cin >> n >>k ;
    if(abs(n-k)/k==0){
        if(k == 1){
        cout << 0 << endl;
        return 0;
    }
     REP(i ,INF){
         ll n_ = min(n,abs(n-k));
         if(n_ == n){
             cout << n << endl;
             return 0;
         }
         n = n_;
          
     }
    }
     else{
         if(n>k){
         ll n1;
         n1 = n-k*(floor(double(n-k)/k)-1);
         cout << min(n1,abs(n1-k)) << endl;
     }
     else{
         ll n1;
         n1 =k-n*(floor(double(k-n)/n)-1);
         cout << min(n1,abs(n1-n))<< endl;

     }
     }

}