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;

#define deb(x) cout << #x << " " << x << endl;
#define fo(i,n) for(int i=0;i<n;i++)
#define Fo(i,k,n) for(int i=k;i<n;i++)
#define vi vector<int>
#define ii pair<int,int>
#define vii vector<ii>
#define ll long long
#define pb push_back
#define endl "\n"
#define mp map<int,int>
#define F first
#define S second
#define sz(v) (int)v.size()
#define mod 1000000007

void c_p_c()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

}

int main(){
	c_p_c();
	ll n,k; cin >> n >> k;
	if(n%k==0){
		cout << 0 << endl;
		return 0;
	}
	if(n-k<0){
		ll mxn = abs(n-k) > n ? n : abs(n-k);
		cout << mxn << endl;
		return 0;
	}

	ll a = n%k;
	ll b = abs((n%k)-k);

	cout << min(a,b) << endl;

	return 0;
}