Problem: One day Polycarpus stopped by a supermarket on his way home. It turns out that the supermarket is having a special offer for stools. The offer is as follows: if a customer's shopping cart contains at least one stool, the customer gets a 50% discount on the cheapest item in the cart (that is, it becomes two times cheaper). If there are several items with the same minimum price, the discount is available for only one of them!

Polycarpus has k carts, and he wants to buy up all stools and pencils from the supermarket. Help him distribute the stools and the pencils among the shopping carts, so that the items' total price (including the discounts) is the least possible.

Polycarpus must use all k carts to purchase the items, no shopping cart can remain empty. Each shopping cart can contain an arbitrary number of stools and/or pencils.

Input

The first input line contains two integers n and k (1 ≤ k ≤ n ≤ 103) — the number of items in the supermarket and the number of carts, correspondingly. Next n lines describe the items as "ci ti" (without the quotes), where ci (1 ≤ ci ≤ 109) is an integer denoting the price of the i-th item, ti (1 ≤ ti ≤ 2) is an integer representing the type of item i (1 for a stool and 2 for a pencil). The numbers in the lines are separated by single spaces.

Output

In the first line print a single real number with exactly one  decimal place — the minimum total price of the items, including the discounts.

In the following k lines print the descriptions of the items in the carts. In the i-th line print the description of the i-th cart as "t b1 b2 ... bt" (without the quotes), where t is the number of items in the i-th cart, and the sequence b1, b2, ..., bt (1 ≤ bj ≤ n) gives the indices of items to put in this cart in the optimal distribution. All indices of items in all carts should be pairwise different, each item must belong to exactly one cart. You can print the items in carts and the carts themselves in any order. The items are numbered from 1 to n in the order in which they are specified in the input.

If there are multiple optimal distributions, you are allowed to print any of them.

Examples

Input

3 2
2 1
3 2
3 1


Output

5.5
2 1 2
1 3


Input

4 3
4 1
1 2
2 2
3 2


Output

8.0
1 1
2 4 2
1 3

Note

In the first sample case the first cart should contain the 1st and 2nd items, and the second cart should contain the 3rd item. This way each cart has a stool and each cart has a 50% discount for the cheapest item. The total price of all items will be: 2·0.5 + (3 + 3·0.5) = 1 + 4.5 = 5.5.

Solution (in JAVA):

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


public class B {
	
	public static void main(String[] args) throws IOException 
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt(), k = sc.nextInt();
		Item[] items = new Item[n];
		for(int i = 0; i < n; ++i)
			items[i] = new Item(sc.nextInt(), sc.nextInt(), i + 1);
		Arrays.sort(items);
		double ans = 0;
		int min = (int)1e9;
		ArrayList<Integer>[] carts = new ArrayList[k];
		for(int i = 0; i < k; ++i)
			carts[i] = new ArrayList<>();
		boolean minimize = false;
		for(int i = 0, curCart = 0; i < n; ++i)
		{
			min = Math.min(min, items[i].cost);
			if(items[i].type == 1 && curCart < k - 1)
				ans += items[i].cost / 2.0;
			else
			{
				minimize |= items[i].type == 1;
				ans += items[i].cost;
			}
			carts[curCart].add(items[i].idx);
			if(curCart < k - 1)
				curCart++;
		}
		if(minimize)
			ans -= min / 2.0;
		out.printf("%.1f\n", ans);
		for(int i = 0; i < k; ++i)
		{
			out.print(carts[i].size());
			for(int x: carts[i])
				out.print(" " + x);
			out.println();
		}
		out.flush();
		out.close();

	}
	
	static class Item implements Comparable<Item>
	{
		int cost, type, idx;
		
		Item(int a, int b, int c) { cost = a; type = b; idx = c; }
		
		public int compareTo(Item i)
		{
			if(type != i.type)
				return type - i.type;
			return i.cost - cost;
		}
	}

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

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

		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
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}

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


	}
}