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)

// SeeAlso: 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define MAX 100000
#define NIL -1
#define MOD 1000000007

typedef int _loop_int;
#define rep(i,n) for(int i = 0; i < n; i++)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1;i>=(_loop_int)(a);--i)

#define debug(x) cout<<#x<<": "<<x<<endl
#define debig_vec(v) cout<<#v<<":";rep(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

// 最大公約数
inline constexpr ll gcd(ll a,ll b){if(!a||!b)return 0;while(b){ll c=b;b=a%b;a=c;}return a;}
// 最小公倍数
inline constexpr ll lcm(ll a,ll b){if(!a||!b)return 0;return a*b/gcd(a,b);}

#define print2D(h, w, arr) rep(i, h) { rep(j, w) cout << arr[i][j] << " "; cout << endl; }
template<class T> void print(const T& x){cout << setprecision(12) << x << endl;}
template<class T, class... A> void print(const T& first, const A&... rest) { cout << first << " "; print(rest...); }

int main() {
    ll n,k;
    cin>>n>>k;
    ll t = n%k;
    print(min(abs(t-k), t));
    return 0;
}