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 (GCC 5.4.1) (judge verdict: Accepted)


#define _USE_MATH_DEFINES
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>  
#include <stdbool.h>   
#define inf (INT_MAX-1)
#define INF 9223372036854775807
#define PI 3.14159265358979323846;
#define EPS 1e-10
#define sq(n) ((n)*(n))
#define rep(i,n) for(i=0;i<n;i++)
#define rev(i,n) for(i=n-1;i>=0;i--)
/*
#define sort(a,n) qsort(a,n,sizeof(TYPE),cmp)
#define sort_r(a,n) qsort(a,n,sizeof(TYPE),cmp_r)
*/
#define chsort(s,n) qsort(s,n,sizeof(char),cmp)
#define chsort_r(s,n) qsort(s,n,sizeof(char),char_cmp_r)
#define TYPE long long
#define ll long long
#define MEMSET(a) memset(a,0,sizeof(a))
#define MEMSET_U(a) memset(a,-1,sizeof(a))
#define bool    _Bool
const int mod = (int)1e09 + 7;

//#define DEBUG1
//#define DEBUG2
//#define DEBUGF
#define DUMMY

int in(void) { int i; scanf("%d", &i); return i; }
long long llin(void) { long long i; scanf("%lld", &i); return i; }
double din(void) { double i; scanf("%lf", &i); return i; }
void chin(char s[]) { scanf("%s", s); }

void print(int a) { printf("%d\n", a); }
void llprint(long long a) { printf("%lld\n", a); }
void dprint(double a) { printf("%.10f\n", a); }
void print2(int a, int b) { printf("%d %d\n", a, b); }
int Max(int a, int b) { if (a > b) { return a; }return b; }
int Min(int a, int b) { if (a < b) { return a; }return b; }
long long llmax(long long a, long long b) { return a > b ? a : b; }
long long llmin(long long a, long long b) { return a < b ? a : b; }
double dmax(double a, double b) { return a > b ? a : b; }
double dmin(double a, double b) { return a < b ? a : b; }
//long long llmax(long long a, long long b) { return a > b ? a : b; }
//long long llmin(long long a, long long b) { return a < b ? a : b; }
//double dmax(double a, double b) { return a > b ? a : b; }
int cmp(const void *a, const void *b) { return *(TYPE *)a - *(TYPE *)b; }
int cmp_r(const void *a, const void *b) { return *(TYPE *)b - *(TYPE *)a; }
int char_cmp(const void *a, const void *b) { return strcmp((char *)a, (char *)b); }
int char_cmp_r(const void *a, const void *b) { return strcmp((char *)b, (char *)a); }
void swap(int *a, int *b) { int t = *a; *a = *b; *b = t; }



int main() {
	int N, K;
	scanf("%d%d", &N, &K);
	int *X = (int *)malloc(sizeof(int)*N);
	for (int i = 0; i < N; i++) {
		scanf("%d", &X[i]);
	}
	int sum = 0;
	for (int i = 0; i < N; i++) {
		if (X[i] < (K - X[i])) {
			sum += X[i]*2;
		}
		else {
			sum += (K - X[i]) * 2;
		}
	}
	int ans = sum;
	printf("%d\n", ans);





#ifdef DEBUGF
	getch();
#endif
	return 0;
}