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
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.text.*;

public class cf312C {
    
    static BufferedReader br;
    static Scanner sc;
    static PrintWriter out;
    
    public static void initA() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            //br = new BufferedReader(new FileReader("input.txt"));
            sc = new Scanner(System.in);
            //out = new PrintWriter("output.txt");
            out = new PrintWriter(System.out);
        } catch (Exception e) {
        }
    }

    static boolean next_permutation(Integer[] p) {
        for (int a = p.length - 2; a >= 0; --a) {
            if (p[a] < p[a + 1]) {
                for (int b = p.length - 1;; --b) {
                    if (p[b] > p[a]) {
                        int t = p[a];
                        p[a] = p[b];
                        p[b] = t;
                        for (++a, b = p.length - 1; a < b; ++a, --b) {
                            t = p[a];
                            p[a] = p[b];
                            p[b] = t;
                        }
                        return true;
                    }
                }
            }
        }
        return false;
    }
    
    public static void initB() {
        try {
            
            br = new BufferedReader(new FileReader("input.txt"));
            sc = new Scanner(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
            
        } catch (Exception e) {
        }
    }
    
    public static String getString() {
        try {
            return br.readLine();
        } catch (Exception e) {
        }
        return "";
    }
    
    public static Integer getInt() {
        try {
            return Integer.parseInt(br.readLine());
        } catch (Exception e) {
        }
        return 0;
    }
    
    public static Integer[] getIntArr() {
        try {
            StringTokenizer temp = new StringTokenizer(br.readLine());
            int n = temp.countTokens();
            Integer temp2[] = new Integer[n];
            for (int i = 0; i < n; i++) {
                temp2[i] = Integer.parseInt(temp.nextToken());
            }
            return temp2;
        } catch (Exception e) {
        }
        return null;
    }
    
    public static Long[] getLongArr() {
        try {
            StringTokenizer temp = new StringTokenizer(br.readLine());
            int n = temp.countTokens();
            Long temp2[] = new Long[n];
            for (int i = 0; i < n; i++) {
                temp2[i] = Long.parseLong(temp.nextToken());
            }
            return temp2;
        } catch (Exception e) {
        }
        return null;
    }
    
    public static String[] getStringArr() {
        try {
            StringTokenizer temp = new StringTokenizer(br.readLine());
            int n = temp.countTokens();
            String temp2[] = new String[n];
            for (int i = 0; i < n; i++) {
                temp2[i] = (temp.nextToken());
            }
            return temp2;
        } catch (Exception e) {
        }
        return null;
    }
    
    public static int getMax(Integer[] ar) {
        int t = ar[0];
        for (int i = 0; i < ar.length; i++) {
            if (ar[i] > t) {
                t = ar[i];
            }
        }
        return t;
    }
    
    public static void print(Object a) {
        out.println(a);
    }
    
    public static void print(String s, Object... a) {
        out.printf(s, a);
    }
    
    public static int nextInt() {
        return sc.nextInt();
    }
    
    public static double nextDouble() {
        return sc.nextDouble();
    }
    
    public static void main(String[] ar) {
        initA();
        solve();
        out.flush();
    }
    static long hitung(long l){
        long byk_el = l-1;
        if(byk_el%2==0){
            return (byk_el+1) * (byk_el/2);
        }else{
            return ((byk_el+1)/2) * (byk_el);
        }
    }
    
    public static void solve() {
        Long xx[] = getLongArr();
        
        long n = xx[0], tle = xx[1];
        long milyar = 1000000000l;
        
        long maksimal = hitung(n);
        
        if(maksimal<=tle){
            //print(maksimal);
            print("no solution");
            return;
        }
        long needed ;
        for(int i =1 ;i <=n ;i++){
            print(i+" "+i*5000);
        }
        
        
    }
}
