Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.

The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.

The pseudo code of the unexpected code is as follows:
    
    
      
    input n  
    for i from 1 to n  
        input the i-th point's coordinates into p[i]  
    sort array p[] by increasing of x coordinate first and increasing of y coordinate second  
    d=INF        //here INF is a number big enough  
    tot=0  
    for i from 1 to n  
        for j from (i+1) to n  
            ++tot  
            if (p[j].x-p[i].x>=d) then break    //notice that "break" is only to be  
                                                //out of the loop "for j"  
            d=min(d,distance(p[i],p[j]))  
    output d  
    

Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.

You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?

Input

A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).

Output

If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.

The conditions below must be held:

  * All the points must be distinct. 
  * |xi|, |yi| ≤ 109. 
  * After running the given code, the value of tot should be larger than k. 

Examples

Input

4 3


Output

0 0
0 1
1 0
1 1


Input

2 100


Output

no solution
import java.io.*;
import java.util.*;

public class C
{
	String line;
	StringTokenizer inputParser;
	BufferedReader is;
	FileInputStream fstream;
	DataInputStream in;
	String FInput="";
	
	
	void openInput(String file)
	{

		if(file==null)is = new BufferedReader(new InputStreamReader(System.in));//stdin
		else
		{
			try{
		
				
			fstream = new FileInputStream(file);
			in = new DataInputStream(fstream);
			is = new BufferedReader(new InputStreamReader(in));
			}catch(Exception e)
			{
				System.err.println(e);
			}
		}

	}
	
	void readNextLine()
	{
		try {
			line = is.readLine();
			inputParser = new StringTokenizer(line, " ");
			//System.err.println("Input: " + line);
		} catch (IOException e) {
			System.err.println("Unexpected IO ERROR: " + e);
		}	
		
	}
	
	int NextInt()
	{
		String n = inputParser.nextToken();
		int val = Integer.parseInt(n);
		
		//System.out.println("I read this number: " + val);
		return val;
	}
	
	String NextString()
	{
		String n = inputParser.nextToken();
		return n;
	}
	
	void closeInput()
	{
		try {
			is.close();
		} catch (IOException e) {
			System.err.println("Unexpected IO ERROR: " + e);
		}
			
	}
	
	public void readFInput()
	{
		for(;;)
		{
			try
			{
				readNextLine();
				FInput+=line+" ";
			}
			catch(Exception e)
			{
				break;
			}
		}
		inputParser = new StringTokenizer(FInput, " ");
	}
	
	long NextLong()
    {
            String n = inputParser.nextToken();
            
            long val = Long.parseLong(n);
            
            return val;
    }
	
	public static void main(String [] argv)
	{
		String filePath=null;
		if(argv.length>0)filePath=argv[0];
		new C(filePath);
	}

	boolean [] [] p;
	
	public C(String inputFile)
	{
		openInput(inputFile);
    	readNextLine();

		int n=NextInt(), k=NextInt();
		
		if(n*n-n>k*2)
		{
			for(int i=0; i<n; i++)
				System.out.println(47+" "+i);
		}
		else
			System.out.println("no solution");
		
	
		
        closeInput();	
	}




}

