Problem: There are n students at Berland State University. Every student has two skills, each measured as a number: ai — the programming skill and bi — the sports skill.

It is announced that an Olympiad in programming and sports will be held soon. That's why Berland State University should choose two teams: one to take part in the programming track and one to take part in the sports track.

There should be exactly p students in the programming team and exactly s students in the sports team. A student can't be a member of both teams.

The university management considers that the strength of the university on the Olympiad is equal to the sum of two values: the programming team strength and the sports team strength. The strength of a team is the sum of skills of its members in the corresponding area, so the strength of the programming team is the sum of all ai and the strength of the sports team is the sum of all bi over corresponding team members.

Help Berland State University to compose two teams to maximize the total strength of the university on the Olympiad.

Input

The first line contains three positive integer numbers n, p and s (2 ≤ n ≤ 3000, p + s ≤ n) — the number of students, the size of the programming team and the size of the sports team.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 3000), where ai is the programming skill of the i-th student.

The third line contains n positive integers b1, b2, ..., bn (1 ≤ bi ≤ 3000), where bi is the sports skill of the i-th student.

Output

In the first line, print the the maximum strength of the university on the Olympiad. In the second line, print p numbers — the members of the programming team. In the third line, print s numbers — the members of the sports team.

The students are numbered from 1 to n as they are given in the input. All numbers printed in the second and in the third lines should be distinct and can be printed in arbitrary order.

If there are multiple solutions, print any of them.

Examples

Input

5 2 2
1 3 4 5 2
5 3 2 1 4


Output

18
3 4 
1 5 


Input

4 2 2
10 8 8 3
10 7 9 4


Output

31
1 2 
3 4 


Input

5 3 1
5 2 5 1 7
6 3 1 6 3


Output

23
1 3 5 
4 

Solution (in JAVA):

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

public class I {
	static int inf = (int)1e8;
	static int n;
	static int p;
	static int s;
	static Pair[] arr;
	static int[] sum;
	static int[][] dp;
	static StringBuilder sb1;
	static StringBuilder sb2;
	static int cnt1 = 1;
	static int cnt2 = 1;
	public static void main(String[] args) throws IOException {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw=new PrintWriter(System.out,true);
		
		StringTokenizer st = new StringTokenizer(br.readLine());
		n =Integer.parseInt(st.nextToken());
		p = Integer.parseInt(st.nextToken());
		s = Integer.parseInt(st.nextToken());
		
		dp = new int[n+5][p+5];
		for (int i = 0; i < dp.length; i++) {
			Arrays.fill(dp[i], -1);
		}
		st = new StringTokenizer(br.readLine());
		int[] prog = new int[n];
		for (int i = 0; i < prog.length; i++) 
			prog[i] = Integer.parseInt(st.nextToken());
		
		st = new StringTokenizer(br.readLine());
		int[] spor = new int[n];
		for (int i = 0; i < spor.length; i++)
			spor[i] = Integer.parseInt(st.nextToken());
			
		arr = new Pair[n];
		for (int i = 0; i < spor.length; i++)
			arr[i] = new Pair(i+1, prog[i], spor[i]);
		
		Arrays.sort(arr);
		sum = new int[n+1];
		for (int i = 1; i <= arr.length; i++) {
			sum[i] = sum[i-1] + arr[i-1].b;
		}
		
		sb1 = new StringBuilder("");
		sb2 = new StringBuilder("");
		pw.println(solve(0, 0));
		printSol(0, 0);
		pw.println(sb1);
		pw.println(sb2);
		pw.flush();
		pw.close();
	}
	
	private static void printSol(int ind, int num) 
	{
		if(num == p)
		{
			int have = ind-num;
			int rem = Math.max(0, s-have);
			for (int i = 0; i < rem; i++) {
				sb2.append(arr[i+ind].id);
				if(i!=rem)
					sb2.append(" ");
			}
			return;
		}
		if(ind >= n)
			return;
		
		if(dp[ind][num] == arr[ind].a + dp[ind+1][num+1])
		{
			sb1.append(arr[ind].id);
			if(cnt1++ != p)
				sb1.append(" ");
			printSol(ind+1, num+1);
		}
		else 
			if(dp[ind][num] == arr[ind].b + dp[ind+1][num])
			{
				sb2.append(arr[ind].id);
				if(cnt2++ != s)
					sb2.append(" ");
				printSol(ind+1, num);
			}
			else
			{
				printSol(ind+1, num);
			}
			
		
//		return dp[ind][num] = Math.max(sol1, sol2);
	}

	public static int solve(int ind, int num)
	{
		if(num == p)
		{
			int have = ind-num;
			int rem = Math.max(0, s-have);
			return dp[ind][num] = sum[ind+rem]-sum[ind];
		}
		if(ind >= n)
			return dp[ind][num] = -inf;
		
		if(dp[ind][num] != -1)
			return dp[ind][num];
		int sol1 = arr[ind].a + solve(ind+1, num+1);
		int sol2 = 0;
		if(ind-num < s)
			sol2 = arr[ind].b + solve(ind+1, num);
		else
			sol2 = solve(ind+1, num);
		
		return dp[ind][num] = Math.max(sol1, sol2);
	}
	
	static class Pair implements Comparable<Pair>
	{
		int id, a, b;
		public Pair(int id, int a, int b)
		{
			this.id = id;
			this.a = a;
			this.b = b;
		}
		@Override
		public int compareTo(Pair arg0) {
			return arg0.b - this.b;
		}
		
		public String toString()
		{
			return  a + " " + b;
		}
	}
}