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.util.*;


public class ClosestPair {

public static void main(String[] args)
{
    Scanner s=new Scanner(System.in);
    int n=s.nextInt();
    int k=s.nextInt();
    point[] p=new point[n];
    int c=0;
    for(int i=0;i<n;i++)
    {   point pt = new point();
        pt.x=0;
        pt.y=c;
        c++;
        p[i]=pt;
    }
    
    

    
    Arrays.sort(p,point.ySort);
    
    
    
    double d=Integer.MAX_VALUE;
    int tot=0;
    for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++)
        {
            ++tot;
            if (p[j].x-p[i].x>=d) 
                break;
            d=Math.min(d,distance(p[i],p[j]));
        }
    if(tot<=k)
    System.out.println("no solution");
    else
    for(int i=0;i<n;i++)
    {
        System.out.println(p[i].x+" " + p[i].y);
    }
}
public static double distance(point p1,point p2)
{
    return (Math.sqrt(Math.pow(p1.x-p2.x, 2)+Math.pow(p1.y-p2.y, 2)));
}

static class point implements Comparable<point>
{
    int x,y;
    public point()
    {
        
    }
    public point(int x, int y)
    {
        this.x=x;
        this.y=y;
    }
    
    public int compareTo(point pt) {
         
        return this.x-pt.x;
    }
    public static Comparator<point> ySort =new Comparator<point>()
            {
               public int compare(point p1,point p2) {
                  int x= p1.x-p2.x;
                  if(p1.x!=p2.x)
                      return x;
                  else
                      return p1.y-p2.y;
               }
            };
}

    
}
