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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;

public class C {
    static BufferedReader reader;
    static StringTokenizer tokenizer;
    static PrintWriter writer;

    static int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    static long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    static String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

    public static void main(String[] args) throws IOException {
        reader = new BufferedReader(new InputStreamReader(System.in));
        tokenizer = null;
        writer = new PrintWriter(System.out);
        solve();
        reader.close();
        writer.close();
    }


    private static void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
        if (k >= n*(n-1)/2) {
            writer.println("no solution");
            return;
        }
        writer.println("0 0");
        writer.println("1 1");
        for (int i = 2; i < n; i++) {
            writer.println("1 " + i);
        }
    }
}

