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>
using namespace std;
typedef long long ll;
typedef long double ld;
const int mod=1000000007;
const double PI = 3.14159265359;
#define mxe(v) *max_element(v.begin(),v.end())
#define mne(v) *min_element(v.begin(),v.end())
#define INF 1000000000000
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define FOR(i,n) for(int i=0;i<(n);i++)
#define FORA(i,a,b) for(int i=(a);i<=(b);i++)
#define FORD(i,a,b) for(int i=(a);i>=(b);i--)
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);
#define count_1(n)  __builtin_popcountll(n)
#define parity(n)   __builtin_parityll(n)   //Even parity 0 as output
void show    (int a[], int arraysize) { for (int i = 0; i < arraysize; ++i)  {cout << a[i] << " ";}	cout<<endl;}
int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b);}
void sieve(int n)
{
    bool prime[n+1];
    memset(prime, true, sizeof(prime));

    for (int p=2; p*p<=n; p++)
    {
        if (prime[p] == true)
        {
            for (int i=p*p; i<=n; i += p)
                prime[i] = false;
        }
    }
    for (int p=2; p<=n; p++)
       if (prime[p])
          cout << p << " ";
}
int main()
{FAST
int n,k;
cin>>n>>k;
if(k>=n)
{
	cout<<min(n,abs(n-k));
}
else
{
	cout<<n%k<<endl;
}
return  0;
}