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:
import time,math,bisect,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
    return int(stdin.readline())
def IO(): # to take string input
    return stdin.readline()
def IP(): # to take tuple as input
    return map(int,stdin.readline().split())
def L(): # to take list as input
    return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
    return stdout.write(str(x))
def PI(x,y): # to print tuple separatedly
    return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
    return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
    if a==0:
        return b
    elif b==0:
        return a
    if a>b:
        return gcd(a%b,b)
    else:
        return gcd(a,b%a)
def sieve():
    li=[True]*1000001
    li[0],li[1]=False,False
    for i in range(2,len(li),1):
        if li[i]==True:
            for j in range(i*i,len(li),i):
                li[j]=False
    prime=[]
    for i in range(1000001):
        if li[i]==True:
            prime.append(i)
    return prime
def setBit(n):
    count=0
    while n!=0:
        n=n&(n-1)
        count+=1
    return count
def readTree(v,e): # to read tree
    adj=[set() for i in range(v+1)]
    for i in range(e):
        u1,u2,w=IP()
        adj[u1].add((u2,w))
        adj[u2].add((u1,w))
    return adj
def dfshelper(adj,i,visited):
    nodes=1
    visited[i]=True
    for ele in adj[i]:
        if visited[ele]==False:
            nd=dfshelper(adj,ele,visited)
            nodes+=nd
    return nodes
def dfs(adj,v): # a schema of bfs
    visited=[False]*(v+1)
    li=[]
    for i in range(v):
        if visited[i]==False:
            nodes=dfshelper(adj,i,visited)
            li.append(nodes)

    return li
#####################################################################################
mx=10**9+7
def solve():
    n,k=IP()
    a,b,ab=[],[],[]
    at,bt=0,0
    for i in range(n):
        t=L()
        if t[-2]==1 and t[-1]==1:
            ab.append(t[0])
        elif t[-2]==1 and t[-1]==0:
            a.append(t[0])
        elif t[-2]==0 and t[-1]==1:
            b.append(t[0])
        at+=t[0]
        bt+=t[0]
    at,bt=0,0
    t=0
    a.sort(reverse=True)
    b.sort(reverse=True)
    ab.sort(reverse=True)
    while (at<k or bt<k) and (ab or (a and b)):
        if ab and a and b:
            x,y=ab[-1],a[-1]+b[-1]
            if x<=y:
                t+=ab.pop()
            else:
                t=t+a.pop()+b.pop()
        else:
            if ab:
                t+=ab.pop()
            else:
                t=t+a.pop()+b.pop()
        at+=1
        bt+=1
    if at==k and bt==k:
        print(t)
        return
    else:
        print(-1)
        return

solve()

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