Problem: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.

Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.

There are n books in the family library. The i-th book is described by three integers: t_i — the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).

So they need to choose some books from the given n books in such a way that:

  * Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set; 
  * the total reading time of these books is minimized (they are children and want to play and joy as soon a possible). 



The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.

Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5).

The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 ≤ t_i ≤ 10^4, 0 ≤ a_i, b_i ≤ 1), where:

  * t_i — the amount of time required for reading the i-th book; 
  * a_i equals 1 if Alice likes the i-th book and 0 otherwise; 
  * b_i equals 1 if Bob likes the i-th book and 0 otherwise. 

Output

If there is no solution, print only one integer -1. Otherwise print one integer T — the minimum total reading time of the suitable set of books.

Examples

Input

8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0


Output

18


Input

5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0


Output

8


Input

5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1


Output

-1

Solution (in PYTHON3):

Solution:
# e_part_2_att_1.py
#attempted after contest
import sys

def input():
	return sys.stdin.readline().rstrip()

def input_split():
	return [int(i) for i in input().split()]

def add_book():
	global freqs, cf, cp
	# cost_change = 0
	while(True):
		if cf < freqs[cp]:
			cf += 1
			return cp
		else:
			cp += 1
			cf = 0

def remove_book():
	global freqs, cf, cp
	# cost_change = 0
	while(True):
		if cf > 0:
			cf -= 1
			return (-cp)
		else:
			cp -= 1
			cf = freqs[cp]

def make_available(b_time):
	global freqs, cf, cp
	freqs[b_time] += 1
	if b_time < cp: #had a bug here
		return b_time + remove_book()
	else:
		return 0

def make_unavailable(b_time):
	global freqs, cf, cp
	
	if b_time < cp: #had a bug here
		freqs[b_time] -= 1
		# a book was made unavailable that was being used
		return (-b_time + add_book())
	elif (b_time == cp and cf == freqs[cp]):
		freqs[b_time] -= 1
		cf -= 1		
		return (-b_time + add_book())

	else:
		freqs[b_time] -= 1
		return 0 #no change


n,m, k = input_split()
times = []
alice_likes = []
bob_likes = []
for _ in range(n):
	t, a, b = input_split()
	times.append(t)
	alice_likes.append(a)
	bob_likes.append(b)

# if (sum(alice_likes) < k or sum(bob_likes)< k):
# 	ans = -1
# else:
	#worst case choose all, but possible	
times_both = []
times_alice = []
times_bob = []
times_none = []
# books_alice = []

for book in range(n):
	if alice_likes[book] == 1 and bob_likes[book] == 1:
		times_both.append((times[book], book))
	elif alice_likes[book] == 1:
		times_alice.append((times[book], book))
	elif bob_likes[book] == 1:
		times_bob.append((times[book], book))
	else:
		times_none.append((times[book], book))
		# pass

times_both.sort()
times_alice.sort()
times_bob.sort()

# def split_list(lis):
	# return ([a for (a,b) in lis],[b for (a,b) in lis])

# books_alice = [j for (i,j) in times_alice]
# times_alice = 


started = False
possibilities = []
for i in range(k+1):
	# print('i is {}'.format(i))
	both = i
	indi = (k - i) #decreases by 1 in each iter, frees up two books
	extra = m - (both + 2*indi)
	# print('both {}, indi {}, extra {}, nu	m_both')
	if both > len(times_both) or indi > min(len(times_alice), len(times_bob)) or extra < 0:
		continue

	if not started:
		# print('Enter')
		started = True #when i comment this it works on test 10, so mere else vale mei hi  problem hai
		#initialise
		ans = sum([i for (i,j) in times_both[:both]]) 
		ans += sum( [i for (i,j) in times_alice[:indi]])
		ans += sum([ i for (i,j) in times_bob[:indi]])

		freqs = [0 for i in range(10001)]
		pending = (times_both[both:] + times_bob[indi:] + times_alice[indi:] + times_none)
		for tim, _ in pending:
			freqs[tim] += 1

		cp = 0 #current pointer
		cf = 0 #current fill
		extra_cost = 0
		count = 0
		for _ in range(extra):
			extra_cost += add_book()

		ans += extra_cost
		possibilities.append((ans,both))
	else:
		# print('shouldnt enter here')
		# ans = 0
		#manipulate state
		last_ans = possibilities[-1][0]
		ans = last_ans
		new_book1_time = times_alice[indi][0]
		new_book2_time = times_bob[indi][0]
		old_book_time = times_both[both-1][0]
		ans += old_book_time
		ans -= new_book1_time
		ans -= new_book2_time
		ans += make_available(new_book1_time)
		ans += make_available(new_book2_time)
		ans += make_unavailable(old_book_time)
		ans += add_book()
		possibilities.append((ans,both))
# times_both = times_both + [100000]*(n- len(times_both))
# times_alice = times_alice + [100000]*(n- len(times_alice))
# times_bob = times_bob + [100000]*(n- len(times_bob))

# ans = 0
# p1, p2, p3 = 0, 0, 0
# for i in range(k):
# 	if (times_both[p1] <= times_alice[p2] + times_bob[p3]):
# 		ans += times_both[p1]
# 		p1 += 1
# 	else:
# 		ans += times_alice[p2] + times_bob[p3]
# 		p2 += 1
# 		p3 += 1

# times_bob

if len(possibilities) == 0:
	print(-1)
else:
	ans , both = min(possibilities)
	# picked = [False for i in range(n)]
	# for b in range(n):
	indi = k - both
	extra = m - (indi*2 + both)
	books = []
	books += [j for (i,j) in times_both[:both]]
	books += [j for (i,j) in times_alice[:indi]]
	books += [j for (i,j) in times_bob[:indi]]
	pending = (times_both[both:] + times_bob[indi:] + times_alice[indi:] + times_none)
	pending.sort()
	books += [j for (i,j) in pending[:extra]]
	books = [b+1 for b in books]
	print(ans)
	print(*books, sep = ' ')
	#find the configuration and redo the stupid calculation
	#do 4 passes over all books and get the ans