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 <algorithm>
#include <bitset>
#include <boost/multiprecision/cpp_int.hpp>
#include <algorithm>
#include <climits>
#include <cmath>
#include <cstring>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <set>
#include <numeric>
#include <sstream>
#include <stdio.h>
#include <string>
#include <tuple>
#include <vector>
#include<iomanip>
namespace mp = boost::multiprecision;
using namespace std;
using ull = unsigned long long;
using ld = long double;
using lll = boost::multiprecision::cpp_int;
using ll = long long;
#define FOR(i, N) for(ll i = 0; i < N; ++i)

#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
// using int = ll;
template <typename T>
T gcd(T m, T n) {
    if(n == 0)
        return abs(m);
    return (gcd(n, m % n));
}
void putYN(bool b) {
    if(b) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
}
ll combi(ll n, ll k) {
    ll r = 1, rr = 1;
    for(ll i = 0; i < k; ++i) {
        r *= (n - i);
        rr *= (i + 1);
         r %= 1000000007;
         rr %= 1000000007;
    }
    r /= rr;
     r %= 1000000007;
    return r;
}
#define rep(a,b,c) for(ll a = b; a < c; ++i)
ll mod = 1000000007;
//---------------------------------------------------------------------------------------------------
ll f[101010];
void init() {
    f[0] = 1;
    rep(i, 1, 101010) f[i] = (f[i - 1] * i) % mod;
}
//---------------------------------------------------------------------------------------------------
ll inv(ll x) {
    ll res = 1;
    ll k = mod - 2;
    ll y = x;
    while (k) {
        if (k & 1) res = (res * y) % mod;
        y = (y * y) % mod;
        k /= 2;
    }
    return res;
}
//---------------------------------------------------------------------------------------------------
ll C(int n, int k) {
    ll a = f[n]; // = n!
    ll b = f[n-k]; // = (n-k)!
    ll c = f[k]; // = k!

    ll bc = (b * c) % mod;

    return (a * inv(bc)) % mod;
}
ll pow2_(ll n){
  
  ll div = 1000000007ll;
  ll ans = 0,k=2;
  {
    bitset<48> b(n);
    FOR(i,32){
      ans+=b[i]*k%div;
      ans%=div;
      k*=k;
      k%=div;
    }
  }
  return ans;
}
int main(){
  ll N,K;
  cin>>N>>K;
  cout<<min(N%K,abs(N%K-K))<<endl;
  return 0;
}