Problem: You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.

One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.

If there exists no such matrix then output -1.

Input

The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).

Output

If the answer exists then output resulting matrix. Otherwise output -1.

Examples

Input

2 1


Output

1 0 
0 0 


Input

3 2


Output

1 0 0 
0 1 0 
0 0 0 


Input

2 5


Output

-1

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
 
 
public class MaximalBinaryMatrix{
   public static void main(String[] args) {
      MyScanner sc = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      
      int n = sc.nextInt();
      int k = sc.nextInt();
      int[] diagonal = new int[n];
      if(k > n*n){
    	  out.print(-1);
    	  out.close();
    	  return;
      }
      for(int i = n; i > 0; --i){
    	  int tmp = 2*i-1;
    	  if(tmp <= k){
    		  k -= tmp;
    		  diagonal[0] = i;
    		  break;
    	  }
      }
      for(int i = 1; i < n; ++i){
    	for(int j = diagonal[i-1]; j > 0; --j){
    		if(j > n-i) continue;
    		int tmp = 2*j-1;
    		if(tmp <= k){
    			k -= tmp;
    			diagonal[i] = j;
    			break;
    		}
    	}
      }
      for(int i = 0; i < n; ++i){
    	  for(int j = 0; j < n; ++j){
    		  int tmpMin = j, tmpMax = i;
    		  if(j > i){
    			  tmpMax = j;
    			  tmpMin = i;
    		  }
    		  tmpMax -= tmpMin;
    		  out.print(((tmpMax < diagonal[tmpMin]) ? "1" : "0") + " ");
    	  }
    	  out.println();
      }
      
      // Start writing your solution here. -------------------------------------
   
      /*
      int n      = sc.nextInt();        // read input as integer
      long k     = sc.nextLong();       // read input as long
      double d   = sc.nextDouble();     // read input as double
      String str = sc.next();           // read input as String
      String s   = sc.nextLine();       // read whole line as String

      int result = 3*n;
      out.println(result);                    // print via PrintWriter
      */

      // Stop writing your solution here. -------------------------------------
      out.close();
   }

     

   public static PrintWriter out;
      
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
 
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
 
      int nextInt() {
          return Integer.parseInt(next());
      }
 
      long nextLong() {
          return Long.parseLong(next());
      }
 
      double nextDouble() {
          return Double.parseDouble(next());
      }
 
      String nextLine(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }

   }
 
}