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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            if (k >= (n * (n - 1)) / 2) {
                out.println("no solution");
                return;
            }
            for (int i = 0; i < n; i++) {
                out.println(0 + " " + i);
            }
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

