Problem: A student's life is fraught with complications. Some Berland University students know this only too well. Having studied for two years, they contracted strong antipathy towards the chairperson of some department. Indeed, the person in question wasn't the kindest of ladies to begin with: prone to reforming groups, banning automatic passes and other mean deeds. At last the students decided that she just can't get away with all this anymore...

The students pulled some strings on the higher levels and learned that the next University directors' meeting is going to discuss n orders about the chairperson and accept exactly p of them. There are two values assigned to each order: ai is the number of the chairperson's hairs that turn grey if she obeys the order and bi — the displeasement of the directors if the order isn't obeyed. The students may make the directors pass any p orders chosen by them. The students know that the chairperson will obey exactly k out of these p orders. She will pick the orders to obey in the way that minimizes first, the directors' displeasement and second, the number of hairs on her head that turn grey.

The students want to choose p orders in the way that maximizes the number of hairs on the chairperson's head that turn grey. If there are multiple ways to accept the orders, then the students are keen on maximizing the directors' displeasement with the chairperson's actions. Help them.

Input

The first line contains three integers n (1 ≤ n ≤ 105), p (1 ≤ p ≤ n), k (1 ≤ k ≤ p) — the number of orders the directors are going to discuss, the number of orders to pass and the number of orders to be obeyed by the chairperson, correspondingly. Each of the following n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 109), describing the corresponding order.

Output

Print in an arbitrary order p distinct integers — the numbers of the orders to accept so that the students could carry out the revenge. The orders are indexed from 1 to n in the order they occur in the input. If there are multiple solutions, you can print any of them.

Examples

Input

5 3 2
5 6
5 8
1 3
4 3
4 11


Output

3 1 2 

Input

5 3 3
10 18
18 17
10 20
20 18
20 18


Output

2 4 5 

Note

In the first sample one of optimal solutions is to pass orders 1, 2, 3. In this case the chairperson obeys orders number 1 and 2. She gets 10 new grey hairs in the head and the directors' displeasement will equal 3. Note that the same result can be achieved with order 4 instead of order 3.

In the second sample, the chairperson can obey all the orders, so the best strategy for the students is to pick the orders with the maximum sum of ai values. The chairperson gets 58 new gray hairs and the directors' displeasement will equal 0.

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import static java.lang.Math.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.StringTokenizer;
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 * 332C - Students' Revenge

 Let’s sort orders ascending bi, and by equality of bi — descending ai. One can assume that in an optimal solution all the orders obeyed by the chairperson go in the sorted list after orders that she hasn’t obeyed (it may be wrong if there are several same orders, but it doesn’t affect parameters of an answer). Let’s iterate through i — the position of the first order in the sorted list, which the chairperson will obey. To the left of this order we should choose p - k orders which the chairperson won’t obey. As we should choose orders with the maximum sum of bi, we can just choose p - k orders that immediately precede the i-th order. To the right of the i-th order we should choose k - 1 orders which the chairperson will obey. These orders should have the maximum sum of ai. If we iterate i by descending, we can keep these k - 1 orders in some data structure that can perform basic operations with sets in logarithmic time (for example, multiset in C++).

 Asymptotics of the solution — O(nlogn)
 */

/**
 *
 * @author pttrung
 */
public class C {

    public static long Mod = 1000000009;

    public static void main(String[] args) {
        Scanner in = new Scanner();
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int p = in.nextInt();
        int k = in.nextInt();
        Edge[] orders = new Edge[n];
        for (int i = 0; i < n; i++) {
            orders[i] = new Edge(in.nextInt(), in.nextInt(), i + 1);
        }
        Arrays.sort(orders, new Comparator<Edge>() {
            @Override
            public int compare(Edge o1, Edge o2) {
                if (o1.b != o2.b) {
                    return o1.b - o2.b;
                } else {
                    return o2.a - o1.a;
                }
            }
        });
        for (int i = 0; i < n; i++) {
            orders[i].num = i;
        }
        int start = p - k;
        Edge[] second = Arrays.copyOfRange(orders, p - k, n);
        Arrays.sort(second, new Comparator<Edge>() {
            @Override
            public int compare(Edge o1, Edge o2) {
                if (o1.a != o2.a) {
                    return o1.a - o2.a;
                }
                if (o1.num < o2.num) {
                    return -1;
                }
                return 1;
            }
        });
        int minNum = n + 1;
        for(int i = second.length - 1; i >= second.length - k; i--){
            out.print(second[i].index + " ");
            minNum = Math.min(minNum , second[i].num);
        }
        minNum--;
        for(int i = 0; i < p - k ; i++){
            out.print(orders[minNum - i].index + " ");
        }

        out.close();
    }

    static class Edge implements Comparable<Edge> {

        int a, b, index, num;

        public Edge(int a, int b, int index) {
            this.a = a;
            this.b = b;
            this.index = index;
        }

        @Override
        public int compareTo(Edge o) {
            return b - o.b;
        }
    }

    static void check(Point a, Point b, ArrayList<Point> p, Point[] rec, int index) {
        for (int i = 0; i < 4; i++) {
            int m = (i + index) % 4;
            int j = (i + 1 + index) % 4;
            Point k = intersect(minus(b, a), minus(rec[m], rec[j]), minus(rec[m], a));
            if (k.x >= 0 && k.x <= 1 && k.y >= 0 && k.y <= 1) {
                Point val = new Point(k.x * minus(b, a).x, k.x * minus(b, a).y);
                p.add(add(val, a));
                // System.out.println(a + " " + b + " " + rec[i] + " " + rec[j]);
                // System.out.println(add(val, a));
            }
        }
    }

    static Point intersect(Point a, Point b, Point c) {
        double D = cross(a, b);
        if (D != 0) {
            return new Point(cross(c, b) / D, cross(a, c) / D);
        }
        return null;
    }

    static Point convert(Point a, double angle) {
        double x = a.x * cos(angle) - a.y * sin(angle);
        double y = a.x * sin(angle) + a.y * cos(angle);
        return new Point(x, y);
    }

    static Point minus(Point a, Point b) {
        return new Point(a.x - b.x, a.y - b.y);
    }

    static Point add(Point a, Point b) {
        return new Point(a.x + b.x, a.y + b.y);
    }

    static double cross(Point a, Point b) {
        return a.x * b.y - a.y * b.x;
    }

    static class Point {

        double x, y;

        Point(double x, double y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public String toString() {
            return "Point: " + x + " " + y;
        }
    }

    static class Scanner {

        BufferedReader br;
        StringTokenizer st;

        public Scanner() {
            // System.setOut(new PrintStream(new BufferedOutputStream(System.out), true));
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() {

            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    throw new RuntimeException();
                }
            }
            return st.nextToken();
        }

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

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

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

        public String nextLine() {
            st = null;
            try {
                return br.readLine();
            } catch (Exception e) {
                throw new RuntimeException();
            }
        }

        public boolean endLine() {
            try {
                String next = br.readLine();
                while (next != null && next.trim().isEmpty()) {
                    next = br.readLine();
                }
                if (next == null) {
                    return true;
                }
                st = new StringTokenizer(next);
                return st.hasMoreTokens();
            } catch (Exception e) {
                throw new RuntimeException();
            }
        }
    }
}