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.io.*;
import java.util.*;

public class A implements Runnable {
	public static void main (String[] args) {new Thread(null, new A(), "_cf", 1 << 28).start();}
	
	public void run() {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		System.err.println("Go!");

		int a = fs.nextInt();
		int b = fs.nextInt();
		boolean swap = false;
		if(a > b) {
			int t = b;
			b = a; a = t;
			swap = true;
		}
		ArrayList<Integer> forA = new ArrayList<Integer>();
		ArrayList<Integer> forB = new ArrayList<Integer>();
		int lo = 0, hi = 100000;
		for(int bs = 0; bs < 50 && lo <= hi; bs++) {
			int mid = (lo + hi) / 2;
			ArrayList<Integer> tA = new ArrayList<>();
			ArrayList<Integer> tB = new ArrayList<>();
			boolean[] used = new boolean[mid + 1];
			int bb = b, aa = a;
			for(int i = mid; i >= 1; i--) {
				if(bb >= i) {
					bb -= i;
					used[i] = true;
					tB.add(i);
				}
			}
			for(int i = mid; i >= 1; i--) if(!used[i]) {
				aa -= i;
				tA.add(i);
			}
			if(aa >= 0) {
				forA = tA; forB = tB;
				lo = mid+1;
			}
			else {
				hi = mid-1;
			}
		}
		if(swap) {
			ArrayList<Integer> t = new ArrayList<>(forA);
			forA = new ArrayList<>(forB);
			forB = t;
		}
		out.println(forA.size());
		for(int i : forA) out.print(i + " ");
		out.println();
		out.println(forB.size());
		for(int i : forB) out.print(i + " ");
		out.println();
		out.close();
	}
	
	class FastScanner {
		public int BS = 1<<16;
		public char NC = (char)0;
		byte[] buf = new byte[BS];
		int bId = 0, size = 0;
		char c = NC;
		double num = 1;
		BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) throws FileNotFoundException {
			in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
		}

		public char nextChar(){
			while(bId==size) {
				try {
					size = in.read(buf);
				}catch(Exception e) {
					return NC;
				}                
				if(size==-1)return NC;
				bId=0;
			}
			return (char)buf[bId++];
		}

		public int nextInt() {
			return (int)nextLong();
		}

		public long nextLong() {
			num=1;
			boolean neg = false;
			if(c==NC)c=nextChar();
			for(;(c<'0' || c>'9'); c = nextChar()) {
				if(c=='-')neg=true;
			}
			long res = 0;
			for(; c>='0' && c <='9'; c=nextChar()) {
				res = (res<<3)+(res<<1)+c-'0';
				num*=10;
			}
			return neg?-res:res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c!='.' ? cur:cur+nextLong()/num;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c>32) {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c!='\n') {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if(c>32)return true;
			while(true) {
				c=nextChar();
				if(c==NC)return false;
				else if(c>32)return true;
			}
		}
		
		public int[] nextIntArray(int n) {
			int[] res = new int[n];
			for(int i = 0; i < n; i++) res[i] = nextInt();
			return res;
		}
		
	}

	
}