Problem: In a galaxy far, far away Lesha the student has just got to know that he has an exam in two days. As always, he hasn't attended any single class during the previous year, so he decided to spend the remaining time wisely.

Lesha knows that today he can study for at most a hours, and he will have b hours to study tomorrow. Note that it is possible that on his planet there are more hours in a day than on Earth. Lesha knows that the quality of his knowledge will only depend on the number of lecture notes he will read. He has access to an infinite number of notes that are enumerated with positive integers, but he knows that he can read the first note in one hour, the second note in two hours and so on. In other words, Lesha can read the note with number k in k hours. Lesha can read the notes in arbitrary order, however, he can't start reading a note in the first day and finish its reading in the second day.

Thus, the student has to fully read several lecture notes today, spending at most a hours in total, and fully read several lecture notes tomorrow, spending at most b hours in total. What is the maximum number of notes Lesha can read in the remaining time? Which notes should he read in the first day, and which — in the second?

Input

The only line of input contains two integers a and b (0 ≤ a, b ≤ 10^{9}) — the number of hours Lesha has today and the number of hours Lesha has tomorrow.

Output

In the first line print a single integer n (0 ≤ n ≤ a) — the number of lecture notes Lesha has to read in the first day. In the second line print n distinct integers p_1, p_2, …, p_n (1 ≤ p_i ≤ a), the sum of all p_i should not exceed a.

In the third line print a single integer m (0 ≤ m ≤ b) — the number of lecture notes Lesha has to read in the second day. In the fourth line print m distinct integers q_1, q_2, …, q_m (1 ≤ q_i ≤ b), the sum of all q_i should not exceed b.

All integers p_i and q_i should be distinct. The sum n + m should be largest possible.

Examples

Input

3 3


Output

1
3 
2
2 1 

Input

9 12


Output

2
3 6
4
1 2 4 5

Note

In the first example Lesha can read the third note in 3 hours in the first day, and the first and the second notes in one and two hours correspondingly in the second day, spending 3 hours as well. Note that Lesha can make it the other way round, reading the first and the second notes in the first day and the third note in the second day.

In the second example Lesha should read the third and the sixth notes in the first day, spending 9 hours in total. In the second day Lesha should read the first, second fourth and fifth notes, spending 12 hours in total.

Solution (in JAVA):

Solution:
import java.util.*;

import java.io.*;
import java.math.*;
import java.awt.geom.*;

import static java.lang.Math.*;

public class Solution implements Runnable {

	long mod1 = (long) 1e9 + 7;
	int mod2 = 998244353;

	public void solve() throws Exception {
		long a=sc.nextLong();
		long b=sc.nextLong();
		long sum=a+b;
		long max=0;
		long start=1;
		while(true)
		{
			if(start*(start+1)/2>a+b)
				break;
			start++;
		}
		start--;
		long lower=Math.min(a,b);
		long higher=Math.min(a,b);
		long lc=lower;
		long hc=higher;
		int arr[]=new int[(int)start+1];
		for(int i=1;i<=start;i++)
		{
			if(lc<i)
				break;
			if(lc<=start)
			{
				//out.println(lc);
				arr[(int)lc]=1;
				lc=0;
			}
			else
			{
				arr[i]=1;
				lc-=i;
			}
			if(lc<=0)
				break;
		}
		if(lc!=0)
		{
			//out.println("fuck");
			arr[(int)start]=1;
			long remain=start-lc;
			arr[(int)remain]=0;
		}
		int one=0;
		int two=0;
		long sum1=0;
		long sum2=0;
		
		for(int i=1;i<=start;i++)
		{
			if(arr[i]==1)
			{
				one++;
				sum1+=i;
			}
			else sum2+=i;
		}
		//out.println(sum1+" "+sum2);
		if(lower==a)
		{
			out.println(one);
			for(int i=1;i<=start;i++)
			{
				if(arr[i]!=0)
				{
					out.print(i+" ");
				}
			}
			out.println();
			out.println(start-one);
			for(int i=1;i<=start;i++)
			{
				if(arr[i]==0)
				{
					out.print(i+" ");
				}
			}
		}
		else 
		{
			out.println(start-one);
			for(int i=1;i<=start;i++)
			{
				if(arr[i]==0)
				{
					out.print(i+" ");
				}
			}
			out.println();
			out.println(one);
			for(int i=1;i<=start;i++)
			{
				if(arr[i]!=0)
				{
					out.print(i+" ");
				}
			}
		}
		
		
		
		

	}

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static long ncr(int n, int r, long p) {
		if (r > n)
			return 0l;
		if (r > n - r)
			r = n - r;

		long C[] = new long[r + 1];

		C[0] = 1;

		for (int i = 1; i <= n; i++) {

			for (int j = Math.min(i, r); j > 0; j--)
				C[j] = (C[j] + C[j - 1]) % p;
		}
		return C[r] % p;
	}

	public long power(long x, long y, long p) {
		long res = 1;
		// out.println(x+" "+y);
		x = x % p;
		if (x == 0)
			return 0;

		while (y > 0) {
			if ((y & 1) == 1)
				res = (res * x) % p;
			y = y >> 1;
			x = (x * x) % p;
		}
		return res;
	}

	static Throwable uncaught;

	BufferedReader in;
	FastScanner sc;
	PrintWriter out;

	@Override
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			sc = new FastScanner(in);
			solve();
		} catch (Throwable uncaught) {
			Solution.uncaught = uncaught;
		} finally {
			out.close();
		}
	}

	public static void main(String[] args) throws Throwable {
		Thread thread = new Thread(null, new Solution(), "", (1 << 26));
		thread.start();
		thread.join();
		if (Solution.uncaught != null) {
			throw Solution.uncaught;
		}
	}

}

class FastScanner {

	BufferedReader in;
	StringTokenizer st;

	public FastScanner(BufferedReader in) {
		this.in = in;
	}

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

	public int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}

	public int[] readArray(int n) throws Exception {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long nextLong() throws Exception {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws Exception {
		return Double.parseDouble(nextToken());
	}

}