There are $n$ pairs of hard disk drives (HDDs) in a cluster. Each HDD is located at an integer coordinate on an infinite straight line, and each pair consists of one primary HDD and one backup HDD.

Next, you want to place $\boldsymbol{\mbox{k}}$ computers at integer coordinates on the same infinite straight line. Each pair of HDDs must then be connected to a single computer via wires, but a computer can have any number (even zero) of HDDs connected to it. The length of a wire connecting a single HDD to a computer is the absolute value of the distance between their respective coordinates on the infinite line. We consider the total length of wire used to connect all the HDDs to computers to be the sum of the lengths of all the wires used to connect HDDs to computers. Note that both the primary and secondary HDDs in a pair must connect to the same computer.

Given the locations of $n$ pairs (i.e., primary and backup) of HDDs and the value of $\boldsymbol{\mbox{k}}$, place all $\boldsymbol{\mbox{k}}$ computers in such a way that the total length of wire needed to connect each pair of HDDs to computers is minimal. Then print the total length on a new line.

Input Format

The first line contains two space-separated integers denoting the respective values of $n$ (the number of pairs of HDDs) and $\boldsymbol{\mbox{k}}$ (the number of computers). 

Each line $\boldsymbol{i}$ of the $n$ subsequent lines contains two space-separated integers describing the respective values of $a_i$ (coordinate of the primary HDD) and $b_i$ (coordinate of the backup HDD) for a pair of HDDs.

Constraints

$2\leq k\leq n\leq10^5$  
$4\leq k\times n\leq10^5$  
$-10^9\leq a_i,b_i\leq10^9$  

Output Format

Print a single integer denoting the minimum total length of wire needed to connect all the pairs of HDDs to computers.

Sample Input
5 2
6 7
-1 1
0 1
5 2
7 3

Sample Output
13

Explanation

For the given Sample Case, it's optimal to place computers at positions $\mbox{0}$ and $\boldsymbol{6}$ on our infinite line. We then connect the second ($a=-1,b=1$) and the third ($a=0,b=1$) pairs of HDDs to the first computer (at position $\mbox{0}$) and then connect the remaining pairs to the second computer (at position $\boldsymbol{6}$).    

We calculate the wire lengths needed to connect the drives to each computer. The amount of wire needed to connect the second and third drives to the first computer is $(1+1)+(0+1)=3$, and the amount of wire needed to connect the rest of the drives to the second computer is $(0+1)+(1+4)+(1+3)=10$. When we sum the lengths of wire needed to connect all pairs of drives to the two computers, we get a total length of $3+10=13$. Thus, we print $13$ as our answer.

Difficulty: HARD
Tags: Dynamic programming
Skill types: Dynamic programming

Starter code:

Solutions:
from copy import copy
from operator import itemgetter
ap = []
roots = []
saved_cost = dict()

class CostTree:

	def __init__(self, lower, upper):
		(self.lower, self.upper) = (lower, upper)
		self.points = self.sum = 0
		if self.lower == self.upper:
			self.L_branch = self.R_branch = None
		else:
			m = (self.lower + self.upper) // 2
			self.L_branch = CostTree(lower, m)
			self.R_branch = CostTree(m + 1, upper)

	def update(self, point):
		new = copy(self)
		new.points += 1
		new.sum += ap[point]
		if self.lower != self.upper:
			m = (new.lower + new.upper) // 2
			if point <= m:
				new.L_branch = new.L_branch.update(point)
			else:
				new.R_branch = new.R_branch.update(point)
		return new

def grow_cost_trees(p, n, k):
	all_points = []
	for (i, x) in enumerate(p):
		all_points.extend([(x[0], i), (x[1], i)])
	all_points.sort(key=itemgetter(0))
	(current, count, r) = (None, -1, [[] for i in range(n)])
	for i in range(2 * n):
		if all_points[i][0] != current:
			current = all_points[i][0]
			ap.append(current)
			count += 1
		r[all_points[i][1]].append(count)
	roots.append(CostTree(0, len(ap) - 1))
	for i in range(n):
		roots.append(roots[-1].update(r[i][0]).update(r[i][1]))

def cost(i, j):

	def cost_by_trees(i, j):
		b1 = roots[i]
		b2 = roots[j + 1]
		lower_sum = upper_sum = 0
		count_l = count_r = 0
		while True:
			if b1.L_branch is None:
				break
			if b2.L_branch.points - b1.L_branch.points >= j - i + 1 - count_l:
				upper_sum += b2.R_branch.sum - b1.R_branch.sum
				count_r += b2.R_branch.points - b1.R_branch.points
				(b1, b2) = (b1.L_branch, b2.L_branch)
			else:
				lower_sum += b2.L_branch.sum - b1.L_branch.sum
				count_l += b2.L_branch.points - b1.L_branch.points
				(b1, b2) = (b1.R_branch, b2.R_branch)
		return upper_sum - lower_sum + (count_l - count_r) * ap[b1.lower]
	if (i, j) not in saved_cost:
		saved_cost[i, j] = cost_by_trees(i, j)
	return saved_cost[i, j]

def find_least_cost(n, k):

	def recursive(left, right, bottom, top, comp):
		nonlocal row, next_row
		mid = (left + right) // 2
		for i in range(bottom, min(mid, top) + 1):
			temp = row[i] + cost(i + comp, mid + comp)
			if next_row[mid] is None or temp < next_row[mid]:
				next_row[mid] = temp
				L_limit = U_limit = i
			elif temp == next_row[mid]:
				U_limit = i
		if mid - left > 0:
			recursive(left, mid, bottom, L_limit, comp)
		if right - mid > 1:
			recursive(mid + 1, right, U_limit, top, comp)
		return (L_limit, U_limit)
	width = n - k + 1
	row = [cost(0, i) for i in range(width)]
	next_row = [None] * width
	for i in range(1, k - 1):
		(L_limit, _) = recursive(width - 1, width - 1, 0, width - 1, i)
		if width > 1:
			recursive(0, width - 1, 0, L_limit, i)
		(row, next_row) = (next_row, [None] * width)
	recursive(width - 1, width - 1, 0, width - 1, k - 1)
	return next_row[-1]

def solve():
	(n, k) = [int(i) for i in input().split()]
	point = []
	for _ in range(n):
		(i, j) = [int(i) for i in input().split()]
		point.append(tuple(sorted([i, j])))
	point.sort(key=lambda x: x[0] + x[1])
	grow_cost_trees(point, n, k)
	print(find_least_cost(n, k))
	return
solve()