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 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define REP(i,n) for(int i=0;i<n;i++)
#define RREP(i,n) for(int i=n;i>=0;i--)
#define FOR(i,m,n) for(int i=m;i<n;i++)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
using namespace std;
typedef long long ll;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
const ll INF = 1LL << 60;
//const int INF = 1001001001;

int main(void){
    ll n,k;
    cin >> n >> k;
    ll q=n/k,r=n%k;
    double minn = q+(double)r/(double)k;
    ll ans=INF;
    for(ll i=minn-1;i<=minn+1;i++){
        chmin(ans,abs(k*(q-i)+r));
    }
    cout << ans;
    return 0;
}