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.util.*;

import java.io.*;
import java.text.*;

public class E2_1374 {

	public static int bs(ArrayList<Pair> a, int key) {
		int lo = 0;
		int hi = a.size() - 1;
		int ans = -1;
		while (lo <= hi) {
			int mid = lo + hi >> 1;
			if (a.get(mid).val <= key) {
				ans = mid;
				lo = mid + 1;
			} else {
				hi = mid - 1;
			}
		}
		return ans;
	}

	static class Pair implements Comparable<Pair> {
		int idx, val;

		public Pair(int idx, int val) {
			this.idx = idx;
			this.val = val;
		}

		public int compareTo(Pair p) {
			return val - p.val;
		}
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		ArrayList<Pair>[] arr = new ArrayList[4];
		for (int i = 0; i < arr.length; i++) {
			arr[i] = new ArrayList<Pair>();
			arr[i].add(new Pair(0, 0));
		}
		for (int i = 0; i < n; i++) {
			int t = sc.nextInt();
			int x = sc.nextInt() | (sc.nextInt() << 1);
			arr[x].add(new Pair(i + 1, t));
		}
		for (ArrayList<Pair> al : arr)
			Collections.sort(al);
		int[][] presum = new int[4][];
		int[] ans = { 0, 0, 0, 0, Integer.MAX_VALUE, 0 };
		for (int i = 0; i < presum.length; i++) {

			presum[i] = new int[arr[i].size()];
			if (arr[i].size() == 0)
				continue;
			presum[i][0] = arr[i].get(0).val;
			for (int j = 1; j < presum[i].length; j++) {
				presum[i][j] = presum[i][j - 1] + arr[i].get(j).val;
			}
		}
		for (int d = 0; d < arr[3].size(); d++) {

			int c = Math.max(k - d, 0);

			if (c >= arr[1].size() || c >= arr[2].size())
				continue;
			if (d + 2 * c > m)
				continue;
			int lo = 0;
			int hi = (int) 1e4;
			int[] bsans = null;
			while (lo <= hi) {
				int mid = lo + hi >> 1;
				int idx0 = bs(arr[0], mid);
				int idx1 = Math.max(bs(arr[1], mid), c);
				int idx2 = Math.max(bs(arr[2], mid), c);
				int cand = presum[0][idx0] + presum[1][idx1] + presum[2][idx2] + presum[3][d];
				if (idx0 + idx1 + idx2 + d >= m) {
					bsans = new int[] { idx0, idx1, idx2, d, cand - (idx0 + idx1 + idx2 + d - m) * mid, mid };
					hi = mid - 1;
				} else {
					lo = mid + 1;
				}
			}
			if (bsans != null) {
				if (bsans[4] < ans[4]) {
					ans = bsans;
				}
			}
		}
		if (ans[4] == Integer.MAX_VALUE) {
			pw.println(-1);
		} else {
			pw.println(ans[4]);
			while (ans[0] + ans[1] + ans[2] + ans[3] > m) {
				for (int i = 0; i < 4; i++) {
					if (arr[i].get(ans[i]).val == ans[5]) {
						ans[i]--;
						break;
					}
				}
			}
			for (int i = 0; i < 4; i++) {
				for (int j = 1; j <= ans[i]; j++) {
					pw.print(arr[i].get(j).idx + " ");
				}
			}

		}
		pw.close();

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader r) {
			br = new BufferedReader(r);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < arr.length; i++)
				arr[i] = nextInt();
			return arr;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}
}