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:
from math import inf, log2


class SegmentTree:
    def __init__(self, array, func=max):
        self.n = len(array)
        self.size = 2**(int(log2(self.n-1))+1) if self.n != 1 else 1
        self.func = func
        self.default = 0 if self.func != min else inf
        self.data = [self.default] * (2 * self.size)
        self.process(array)

    def process(self, array):
        self.data[self.size : self.size+self.n] = array
        for i in range(self.size-1, -1, -1):
            self.data[i] = self.func(self.data[2*i], self.data[2*i+1])

    def query(self, alpha, omega):
        """Returns the result of function over the range (inclusive)!"""
        if alpha == omega:
            return self.data[alpha + self.size]
        res = self.default
        alpha += self.size
        omega += self.size + 1
        while alpha < omega:
            if alpha & 1:
                res = self.func(res, self.data[alpha])
                alpha += 1
            if omega & 1:
                omega -= 1
                res = self.func(res, self.data[omega])
            alpha >>= 1
            omega >>= 1
        return res

    def update(self, index, value):
        """Updates the element at index to given value!"""
        index += self.size
        self.data[index] = value
        index >>= 1
        while index:
            self.data[index] = self.func(self.data[2*index], self.data[2*index+1])
            index >>= 1


# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# ------------------- fast io --------------------
from math import gcd, ceil

def pre(s):
    n = len(s)
    pi = [0] * n
    for i in range(1, n):
        j = pi[i - 1]
        while j and s[i] != s[j]:
            j = pi[j - 1]
        if s[i] == s[j]:
            j += 1
        pi[i] = j
    return pi


def prod(a):
    ans = 1
    for each in a:
        ans = (ans * each)
    return ans

def lcm(a, b): return a * b // gcd(a, b)


def binary(x, length=16):
    y = bin(x)[2:]
    return y if len(y) >= length else "0" * (length - len(y)) + y


for _ in range(int(input()) if not True else 1):
    n, k = map(int, input().split())
    #c, d = map(int, input().split())
    both = []
    anb = []
    bna = []
    for i in range(n):
        t, a, b = map(int, input().split())
        if a and b:
            both += [t]
        if a and not b:
            anb += [t]
        if b and not a:
            bna += [t]
    both = sorted(both)
    anb = sorted(anb)
    bna = sorted(bna)

    if not both:
        if min(len(anb), len(bna)) < k:
            print(-1)
        else:
            print(sum(anb[:k]) + sum(bna[:k]))
        continue
    if not anb or not bna:
        if len(both) < k:
            print(-1)
        else:
            print(sum(both[:k]))
        continue

    l1, l2, l3 = len(both), len(anb), len(bna)

    both = SegmentTree(both, func=lambda a,b:a+b)
    anb = SegmentTree(anb, func=lambda a,b:a+b)
    bna = SegmentTree(bna, func=lambda a,b:a+b)

    anss = inf
    for i in range(min(l1+1, k+1)):
        ans = both.query(0, i-1) if i else 0
        if min(l2, l3) >= k-i:
            if i != k:
                ans += anb.query(0, k-i-1) + bna.query(0, k-i-1)
            anss = min(anss, ans)
    print(anss if anss != inf else -1)