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 exactly m 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 exactly m 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 m 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 three integers n, m and k (1 ≤ k ≤ m ≤ 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.

If the solution exists, print T in the first line — the minimum total reading time of the suitable set of books. In the second line print m distinct integers from 1 to n in any order — indices of books which are in the set you found.

If there are several answers, print any of them.

Examples

Input

6 3 1
6 0 0
11 1 0
9 0 1
21 1 1
10 1 0
8 0 1


Output

24
6 5 1 

Input

6 3 2
6 0 0
11 1 0
9 0 1
21 1 1
10 1 0
8 0 1


Output

39
4 6 5 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

/*
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

 */
public class E2 {

	static PriorityQueue<Book> pqA, pqB;
	static long total=0;
	static int aHave=0, bHave=0;
	static int booksUsed=0;
	
	static PriorityQueue<Book> cheapestA, cheapestB, cheapestBoth, cheapestNone, worstBoth, worstA, worstB, worstNone;
	
	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		PrintWriter out=new PrintWriter(System.out);
		int n=fs.nextInt(), readTotal=fs.nextInt(), readEach=fs.nextInt();
		Book[] books=new Book[n], unsorted=new Book[n];
		for (int i=0; i<n; i++) unsorted[i]=books[i]=new Book(fs.nextInt(), fs.nextInt(), fs.nextInt());
		Arrays.sort(books);
		pqA=new PriorityQueue<>((a, b)-> {
			return b.compareTo(a);
		});
		pqB=new PriorityQueue<>((a, b)-> {
			return b.compareTo(a);
		});
		
		cheapestA=new PriorityQueue<>();//untaken
		cheapestB=new PriorityQueue<>();
		cheapestBoth=new PriorityQueue<>();
		cheapestNone=new PriorityQueue<>();
		
		worstBoth=new PriorityQueue<>((a, b)-> {//taken
			return b.compareTo(a);
		});
		worstA=new PriorityQueue<>((a, b)-> {//taken
			return b.compareTo(a);
		});
		worstB=new PriorityQueue<>((a, b)-> {//taken
			return b.compareTo(a);
		});
		worstNone=new PriorityQueue<>((a, b)-> {//taken
			return b.compareTo(a);
		});
		
		
		for (Book b:books) {
			if (b.a && b.b) cheapestBoth.add(b);
			else if (b.a) cheapestA.add(b);
			else if (b.b) cheapestB.add(b);
			else cheapestNone.add(b);
			
			if (!b.a && !b.b) continue;
			if (b.a && b.b) {
				//both
				if (aHave<readEach || bHave<readEach) {
					add(b);
					if (aHave>readEach && !pqA.isEmpty()) {
						Book toRemove=pqA.remove();
						remove(toRemove);
					}
					if (bHave>readEach && !pqB.isEmpty()) {
						Book toRemove=pqB.remove();
						remove(toRemove);
					}
				}
				else {
					if (!pqA.isEmpty() && !pqB.isEmpty()) {
						int bonus=pqA.peek().time+pqB.peek().time;
						if (bonus>b.time) {
							Book removed1=pqA.remove(), removed2=pqB.remove();
							add(b);
							remove(removed1);
							remove(removed2);
						}
					}
				}
			}
			else if (b.a) {
				if (aHave<readEach) {
					add(b);
				}
			}
			else {
				if (bHave<readEach) {
					add(b);
				}
			}
		}
		
		//We might have used too many books
		while (booksUsed>readTotal) {
			//try add 11 back in
			Book toAdd=getCheapestBothUnused();
			Book worstA=getWorstAUsed();
			Book worstB=getWorstBUsed();
			if (toAdd==null || worstA==null || worstB==null) {
				out.println(-1);
				out.close();
				return;
			}
			remove(worstA);
			remove(worstB);
			add(toAdd);
		}
		
		//now we might not have enough books
		while (booksUsed<readTotal) {
			Book cheapestA=getCheapestAUnused();
			Book cheapestB=getCheapestBUnused();
			Book cheapestBoth=getCheapestBothUnused();
			Book cheapestNone=getCheapestNoneUnused();
			
			Book bestToAdd=minCost(cheapestA, cheapestB, cheapestBoth, cheapestNone);
			Book worstBoth=getWorstBothUsed();
			if (worstBoth==null || cheapestA==null || cheapestB==null) {
				//forced to do bestToAdd
				if (bestToAdd==null) {
					out.println(-1);
					out.close();
					return;
				}
				add(bestToAdd);
			}
			else {
				if (bestToAdd==null) {
					//shouldn't even happen
					throw null;
				}
				long costBestToAdd=bestToAdd.time;
				long costSwap=cheapestA.time+cheapestB.time-worstBoth.time;
				if (costBestToAdd<costSwap) {
					add(bestToAdd);
				}
				else {
					remove(worstBoth);
					add(cheapestA);
					add(cheapestB);
				}
			}
		}
		
		
		if (aHave<readEach || bHave<readEach) {
			out.println(-1);
			out.close();
			return;
		}
		else {
			out.println(total);
		}
		
		for (int i=0; i<n; i++) {
			if (unsorted[i].used)
				out.print(i+1+" ");
		}
		out.println();
		out.close();
	}
	
	//TODO: make sure all PQs get updated!!
	static void add(Book b) {
		b.used=true;
		total+=b.time;
		booksUsed++;
		if (b.a && b.b) {
			aHave++;
			bHave++;
			worstBoth.add(b);
		}
		else if(b.a) {
			pqA.add(b);
			worstA.add(b);
			aHave++;
		}
		else if (b.b) {
			pqB.add(b);
			worstB.add(b);
			bHave++;
		}
		else {
			worstNone.add(b);
		}
	}
	
	static void remove(Book b) {
		b.used=false;
		total-=b.time;
		booksUsed--;
		
		if (b.a && b.b) {
			aHave--;
			bHave--;
			cheapestBoth.add(b);
		}
		else if(b.a) {
			aHave--;
			cheapestA.add(b);
		}
		else if (b.b) {
			bHave--;
			cheapestB.add(b);
		}
		else {
			cheapestNone.add(b);
		}
	}
	
	static Book minCost(Book a, Book b, Book c, Book d) {
		Book res=better(a, b);
		res=better(res, c);
		res=better(res, d);
		return res;
	}
	
	static Book better(Book a, Book b) {
		if (a==null) return b;
		if (b==null) return a;
		return a.time<=b.time?a:b;
	}
	
	static Book getCheapestBothUnused() {
		return getCheapestUnused(cheapestBoth);
	}
	
	static Book getCheapestNoneUnused() {
		return getCheapestUnused(cheapestNone);
	}
	
	static Book getCheapestAUnused() {
		return getCheapestUnused(cheapestA);
	}
	
	static Book getCheapestBUnused() {
		return getCheapestUnused(cheapestB);
	}

	static Book getWorstAUsed() {
		return getWorstUsed(worstA);
	}
	
	
	static Book getWorstBUsed() {
		return getWorstUsed(worstB);
	}

	static Book getWorstBothUsed() {
		return getWorstUsed(worstBoth);
	}
	
	static Book getCheapestUnused(PriorityQueue<Book> pq) {
		while (!pq.isEmpty()) {
			Book next=pq.peek();
			if (next.used) pq.remove();
			else return next;
		}
		return null;
	}
	
	static Book getWorstUsed(PriorityQueue<Book> pq) {
		while (!pq.isEmpty()) {
			Book next=pq.peek();
			if (!next.used) pq.remove();
			else return next;
		}
		return null;
	}
	
	static class Book implements Comparable<Book> {
		int time;
		boolean a, b;
		boolean used;
		public Book(int time, int a, int b) {
			this.time=time;
			this.a=a==1;
			this.b=b==1;
		}
		
		public int compareTo(Book o) {
			return Integer.compare(time, o.time);
		}
	}
	
	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

	
}