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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define ll long long 
#define FI first 
#define SE second
#define pb push_back 
#define eb emplace_back
#define mod 1000000007
#define all(c) (c).begin(),(c).end()
#define LB  lower_bound
#define UB  upper_bound
#define max3(a,b,c) max(c,max(a,b))
#define min3(a,b,c) min(c,min(a,b))
#define mems(s, n) memset(s, n, sizeof(s))
#define NINF -1e18
#define INF 1e18
#define int ll int
#define OOK order_of_key  //no of elements less than
#define FBO find_by_order //iterator pointing kth element;indexing starts from 0
#define CK3(x,y,z) cout<<(x)<<" "<<(y)<<" "<<(z)<<endl
#define CK4(a,b,c,d) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<endl

typedef pair<int,int> PII;
typedef pair<pair<int,int>,int> PPII;
typedef pair<int,pair<int,int>> PIPI;
typedef map<int,int>MII;
typedef vector<int> VI;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

/*int power(int a, int b)
{int x=1,y=a;while(b>0){if(b%2){x =(x*y)%mod;}y =(y*y)%mod;b/=2;}return x%mod;}
*/
//int modInverse(int a, int m) {return power(a, m-2);} //gcd(a,m)=1 


signed main()
{
ios::sync_with_stdio(false); cin.tie(0);
int T=1,T1=0;//cin>>T;
while(T1++<T)
{
    int n,k;
    cin>>n>>k;
    int x1=n/k;
    int ans=min(n-x1*k,(x1+1)*k-n);
    cout<<ans<<endl;
}


return 0;
}