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 JAVA):

Solution:


import java.util.*;
import java.io.*;
//Captain on duty!



public class Main {

    static void compare(Main.pair a[], int n) {

        Arrays.sort(a, new Comparator<Main.pair>() {
            @Override
            public int compare(Main.pair p1, Main.pair p2) {
                return p1.f - p2.f;
            }
        });
    }

    public static boolean checkPalindrome(String s) {
        // reverse the given String
        String reverse = new StringBuffer(s).reverse().toString();

        // check whether the string is palindrome or not
        if (s.equals(reverse))
            return true;

        else
            return false;
    }

    static class pair implements Comparable {
        int f;
        int s;

        pair(int fi, int se) {
            f = fi;
            s = se;
        }

        public int compareTo(Object o)//desc order
        {
            pair pr = (pair) o;
            if (s > pr.s)
                return -1;
            if (s == pr.s) {
                if (f > pr.f)
                    return 1;
                else
                    return -1;
            } else
                return 1;
        }

        public boolean equals(Object o) {
            pair ob = (pair) o;
            if (o != null) {
                if ((ob.f == this.f) && (ob.s == this.s))
                    return true;
            }
            return false;
        }

        public int hashCode() {
            return (this.f + " " + this.s).hashCode();
        }


    }

    public static boolean palin(int l, int r, char[] c) {
        while (l <= r) {
            if (c[l] != c[r]) return false;
            l++;
            r--;
        }
        return true;
    }


    public static long gcd(long a, long b) {
        if (b == 0)
            return a;

        return gcd(b, a % b);
    }
    public static long lcm(long a, long b)
    {
        return (a*b)/gcd(a, b);
    }

    public static long hcf(long a, long b) {
        long t;
        while (b != 0) {
            t = b;
            b = a % b;
            a = t;
        }
        return a;
    }

    public static boolean isPrime(long n) {
        if (n <= 1)
            return false;

        // Check from 2 to n-1
        for (int i = 2; i <= Math.sqrt(n) + 1; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    public static String reverse(String str) {
        String str1 = "";
        for (int i = 0; i < str.length(); i++) {
            str1 = str1 + str.charAt(str.length() - i - 1);
        }
        return str1;
    }


    public static double fact(long a) {
        if (a == 1)
            return 1;
        else
            return a * fact(a - 1);
    }

    static boolean isPerfectSquare(double x)
    {

        // Find floating point value of
        // square root of x.
        double sr = Math.sqrt(x);

        // If square root is an integer
        return ((sr - Math.floor(sr)) == 0);
    }


    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }


    }


    public static void main(String[] args) {
        FastReader s = new FastReader();
        //System.out.println("Forces Babyy!!");


        int n=s.nextInt();
        int k=s.nextInt();
        ArrayList both=new ArrayList();
        ArrayList first=new ArrayList();
        ArrayList second =new ArrayList();
        for(int i=0;i<n;i++)
        {
            int t=s.nextInt();
            int a=s.nextInt();
            int b=s.nextInt();
            if(a==1 && b==1)
                both.add(t);
            else if(a==1)
                first.add(t);
            else if(b==1)
                second.add(t);
        }
        Collections.sort(first);
        Collections.sort(second);
        int min=Math.min(first.size(),second.size());
        for(int i=0;i<min;i++)
        {
            both.add((int)first.get(i) + (int)second.get(i));
        }
        Collections.sort(both);
        if(both.size()<k)
            System.out.println(-1);
        else
        {
            long ans=0;
            for(int i=0;i<k;i++)
                ans+=(int)both.get(i);
            System.out.println(ans);
        }







    }

}