Problem:
Merge Sort

Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.

Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1

Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)

Input

In the first line n is given. In the second line, n integers are given.

Output

In the first line, print the sequence S. Two consequtive elements should be separated by a space character.

In the second line, print the number of comparisons.

Constraints

n ≤ 500000

0 ≤ an element in S ≤ 109

Sample Input 1

10
8 5 9 2 6 3 7 1 10 4

Sample Output 1

1 2 3 4 5 6 7 8 9 10
34

Notes

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>


int *WORK;
int compare_count = 0;

int n;

void Merge_sort(int *S, int low, int high)
{
	int mid;
	int i, j, k;


	if(high <= low)
		return;

	mid = (low + high) / 2;

	// 左側の処理
	Merge_sort(S, low, mid);

	// 右側の処理
	Merge_sort(S, mid + 1, high);

	for(i = low; i <= mid; i++)
		WORK[i] = S[i];

	for(i = high, j = mid + 1; i > mid; i--, j++)
		WORK[i] = S[j];

	// マージ
	i = low;
	j = high;
	for(k = low; k <= high; k++){
		if(WORK[i] <= WORK[j])
			S[k] = WORK[i++];
		else
			S[k] = WORK[j--];
		compare_count++;
	}
}


int main()
{
	int *S;
	int i;


	scanf("%d", &n);

	S = (int *)malloc(sizeof(int) * n);
	WORK = (int *)malloc(sizeof(int) * n);

	for(i = 0; i < n; i++)
		scanf("%d", &S[i]);


	Merge_sort(S, 0, n - 1);


	for(i = 0; i < n; i++){
		printf("%d", S[i]);
		if(i != n - 1)
			printf(" ");
		else
			printf("\n");
	}
	printf("%d\n", compare_count);


	return 0;
}