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
# -*- coding: utf-8 -*-
"""
Created on Sat Jul 06 18:16:44 2013

@author: workshop
"""

from __future__ import division;
from bisect import *;
from fractions import Fraction;
import sys;
from math import *;
from fractions import *;
import io;
import re;

INF = 987654321987654321987654321;

def readint(delimiter=' ') :
	return map(int, raw_input().split(delimiter));

def readstr(delimiter=' ') :
	return raw_input().split(delimiter);

def readfloat(delimiter=' ') :
	return map(float, raw_input().split(delimiter));

def index(a, x):
	'Locate the leftmost value exactly equal to x'
	i = bisect_left(a, x)
	if i != len(a) and a[i] == x:
		return i
	raise ValueError

def find_lt(a, x):
	'Find rightmost value less than x'
	i = bisect_left(a, x)
	if i:
		return a[i-1]
	raise ValueError

def find_le(a, x):
	'Find rightmost value less than or equal to x'
	i = bisect_right(a, x)
	if i:
		return a[i-1]
	raise ValueError

def find_gt(a, x):
	'Find leftmost value greater than x'
	i = bisect_right(a, x)
	if i != len(a):
		return a[i]
	raise ValueError

def find_ge(a, x):
	'Find leftmost item greater than or equal to x'
	i = bisect_left(a, x)
	if i != len(a):
		return a[i]
	raise ValueError

def bin_search(a, x, left, right) :

	while left<=right :
		mid = (left + right)//2;
		
		if a[mid] == x :
			return mid;
		elif a[mid] < x :
			left = mid + 1; 
		elif a[mid] > x :
			right = mid - 1;
					   
		pass
	
	return -1;
	pass

def printf(format, *args):
	"""Format args with the first argument as format string, and write.
	Return the last arg, or format itself if there are no args."""
	sys.stdout.write(str(format) % args)
	pass

if __name__ == '__main__':

	n, k = readint();
	
	if (n-1)*(n)/2 <= k :
		print "no solution";
	else :
		for ii in xrange(n) :
			print 0, ii;
			pass
		pass
	
	
	
	pass