Problem:
Score : 200 points

Problem Statement

There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i).
Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N.

In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B.
Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i).
Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N.

When activated, each type of robot will operate as follows.

When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.

When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.

Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.

Constraints

1 \leq N \leq 100

1 \leq K \leq 100

0 < x_i < K

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
K
x_1 x_2 ... x_N

Outputs

Print the minimum possible total distance covered by robots.

Sample Input 1

1
10
2

Sample Output 1

4

There are just one ball, one type-A robot and one type-B robot.

If the type-A robot is used to collect the ball, the distance from the robot to the ball is 2, and the distance from the ball to the original position of the robot is also 2, for a total distance of 4.

Similarly, if the type-B robot is used, the total distance covered will be 16.

Thus, the total distance covered will be minimized when the type-A robot is used. The output should be 4.

Sample Input 2

2
9
3 6

Sample Output 2

12

The total distance covered will be minimized when the first ball is collected by the type-A robot, and the second ball by the type-B robot.

Sample Input 3

5
20
11 12 9 17 12

Sample Output 3

74

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<vector>
#include<string>
#include<math.h>
#include<algorithm>
#include<numeric>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<ULL> VULL;

class MYCP {
public:

	//数値を区切って文字列にする
	static string MakeString_LongLong(vector<long long> const& numbers, string const& str) {
		if (numbers.size() == 0)return "";
		string result = "" + to_string(numbers[0]);
		for (long long i = 1; i < numbers.size(); i++) {
			result += str;
			result += to_string(numbers[i]);
		}

		return result;
	}

	//空白で区切る為のオーバーロード
	static string MakeString_LongLong(vector<long long> const& numbers) {
		if (numbers.size() == 0)return "";
		string result = "" + to_string(numbers[0]);
		for (long long i = 1; i < numbers.size(); i++) {
			result += " ";
			result += to_string(numbers[i]);
		}

		return result;
	}



	//文字列の配列を改行を挟んでまとめる
	static string MakeString_VectorString(vector<string> const& str) {
		string result = "";
		for (long long i = 0; i < str.size(); i++) {
			result += str[i] + "\n";
		}

		return result;
	}

	//文字列を必要な個数だけ読み取る
	static vector<string> MyReadLineSplit(LL n) {
		vector<string> str(n);
		for (long long i = 0; i < n; i++) {
			std::cin >> str[i];
		}

		return str;
	}

	//数値を必要な個数だけ読み取る
	static vector<long long> ReadInts(long long number) {
		vector<long long> a(number);
		for (int i = 0; i < number; i++) {
			std::cin >> a[i];
		}

		return a;
	}

	//渡された自然数が素数ならtureを返す
	static bool PrimeCheck_Int(long long number) {
		if (number < 2)return false;

		for (ULL i = 2; i*i <= number; i++) {
			if (number%i == 0)return false;
		}

		return true;
	}

	//渡された数値以下の素数表を作る
	static vector<long long> MakePrimeList(long long n) {
		vector<long long> list;
		LL i, j, p;
		bool flag;
		for (i = 2; i <= n; i++) {
			flag = true;
			for (j = 0; j < list.size(); j++) {

				if (!(list[j] * list[j] <= i))break;

				if (i%list[j] == 0) {
					flag = false;
					break;
				}
			}
			if (flag)list.push_back(i);
		}

		return list;
	}



	//文字列の分割
	static vector<string> split(string const& str, char sep)
	{
		vector<std::string> v;        // 分割結果を格納するベクター
		auto first = str.begin();              // テキストの最初を指すイテレータ
		while (first != str.end()) {         // テキストが残っている間ループ
			auto last = first;                      // 分割文字列末尾へのイテレータ
			while (last != str.end() && *last != sep)       // 末尾 or セパレータ文字まで進める
				last++;
			v.push_back(string(first, last));       // 分割文字を出力
			if (last != str.end())
				last++;
			first = last;          // 次の処理のためにイテレータを設定
		}
		return v;
	}

	//合計を求める
	static LL Sum(vector<LL> a) {
		LL i, sum = 0;

		for (i = 0; i < a.size(); i++) {
			sum += a[i];
		}

		return sum;
	}

	//小文字ならtrueを返す
	static bool Komoji(char a) {
		if (a >= 'a'&&a <= 'z')return true;

		return false;
	}

	//大文字ならtrueを返す
	static bool Oomoji(char a) {
		if (a >= 'A'&&a <= 'Z')return true;

		return false;
	}

	static LL GreatestCommonFactor(LL a, LL b) {
		LL temp;
		if (a < b) {
			temp = b;
			b = a;
			a = temp;
		}

		while (true)
		{
			temp = a % b;
			a = b;
			b = temp;
			if (b == 0)break;
		}

		return a;
	}

	static LL LeastCommonMultiple(LL a, LL b) {
		return (a / GreatestCommonFactor(a, b))*b;
	}

};

//累積和を求めるクラス
class Syakutori {
private:
	vector<LL> list;

public:

	void MakeArray(vector<LL> data) {
		LL i;
		list = data;
		list.push_back(0);
		list[0] = 0;
		for (i = 1; i < list.size(); i++) {
			list[i] = list[i - 1] + data[i - 1];
		}

	}

	LL Sum(LL start, LL end) {
		if (end < start) {
			cout << "startがendより大きいです";
			return 0;
		}
		if (start < 0 || end >= list.size()) {
			cout << "範囲が異常";
			return 0;
		}

		return list[end] - list[start];
	}




};

string strnum = "0123456789";
string alpS = "abcdefghijklmnopqrstuvwxyz";
string alpL = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

int main(void) {
	int n, k;
	cin >> n >> k;
	vector<int> a(n);
	for (int i = 0; i < n; i++)
		cin >> a[i];
	int sum=0;
	for (int i = 0; i < n; i++) {
		if (a[i] < k - a[i]) {
			sum += a[i];
		}
		else
		{
			sum += k - a[i];
		}
	}
	cout << sum << endl;
	return 0;
}