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)

//Mahir Ratanpara (DA-IICT)
#include<bits/stdc++.h>
using namespace std;
#define Ff(i,a,n) for(i=a;i<n;i++)
#define Fr(i,a,n) for(i=a;i>n;i--)
#define ll long long
#define FAST ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
typedef pair<ll, ll>	pa;
typedef vector<ll>		vc;
typedef vector<pa>		vp;
typedef vector<vc>		vvc;
int mpow(int base, int exp); 
void ipgraph(int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 3e5, M = N;
//=======================

vc g[N];
ll a[N];

int main() {
    FAST;
    ll t, i, j, k, p, q, r, x, y, u, n, m;
    cin>>n>>m;
    cout<<min(n%m,m-n%m);
    
    return 0;
}

int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((ll)result * base) % mod;
    base = ((ll)base * base) % mod;
    exp >>= 1;
  }
  return result;
}

void ipgraph(int n, int m){
	int i, u, v;
	while(m--){
		cin>>u>>v;
		g[u-1].pb(v-1);
		g[v-1].pb(u-1);
	}
}

void dfs(int u, int par){
	for(int v:g[u]){
		if (v == par) continue;
		dfs(v, u);
	}
}