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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
#pragma warning(disable : 4996) //_CRT_SECURE_NO_WARNINGS

int dp[10000][10000];

int comp(const void* a, const void* b) {
 	return  *(int*)a - *(int*)b;
}

int func(long N, long K) {
	/*ベースケース*/
	if (dp[N][K] != -1) return dp[N][K];
	/*再帰処理*/
	if (N - K >= 0) {
		dp[N][K] = N - K;
		func(N - K, K);
	}
	else if (N - K < 0) {
		dp[N][K] = (-1) * (N - K);
		func((-1) * (N - K), K);
	}
}

int main(void) {
	long N, K;
	int res, k = 0;
	int list[10000];
	puts("N,Kを入力してください.\n");
	scanf("%ld%ld", &N, &K);
	/*Kが1の場合最小は0*/
	if (K == 1) {
		printf("%d", 0);
		return 0;
	}
	/*dp初期化*/
	for (int i = 0;i < 10000;i++) {
		for (int j = 0;j < 10000;j++) {
			dp[i][j] = -1;
		}
	}
	/*0回の時の処理*/
	dp[N][0] = N;
	/*再帰関数呼び出し*/
	func(N, K);
	/*-1以外の値を取得し、qsortする。*/
	for (int i = 0;i < 10000;i++) {
		for (int j = 0;j < 10000;j++) {
			if (dp[i][j] != -1) {
				list[k++] = dp[i][j];
			}
		}
	}
	qsort(list, k, sizeof(int), comp);
	printf("%d", list[0]);
	return 0;
}