Problem: This is an interactive problem

You are given a grid n× n, where n is odd. Rows are enumerated from 1 to n from up to down, columns are enumerated from 1 to n from left to right. Cell, standing on the intersection of row x and column y, is denoted by (x, y).

Every cell contains 0 or 1. It is known that the top-left cell contains 1, and the bottom-right cell contains 0.

We want to know numbers in all cells of the grid. To do so we can ask the following questions: 

"? x_1 y_1 x_2 y_2", where 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. In other words, we output two different cells (x_1, y_1), (x_2, y_2) of the grid such that we can get from the first to the second by moving only to the right and down, and they aren't adjacent.

As a response to such question you will be told if there exists a path between (x_1, y_1) and (x_2, y_2), going only to the right or down, numbers in cells of which form a palindrome.

For example, paths, shown in green, are palindromic, so answer for "? 1 1 2 3" and "? 1 2 3 3" would be that there exists such path. However, there is no palindromic path between (1, 1) and (3, 1).

<image>

Determine all cells of the grid by asking not more than n^2 questions. It can be shown that the answer always exists.

Input

The first line contains odd integer (3 ≤ n < 50) — the side of the grid.

Interaction

You begin the interaction by reading n.

To ask a question about cells (x_1, y_1), (x_2, y_2), in a separate line output "? x_1 y_1 x_2 y_2".

Numbers in the query have to satisfy 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. Don't forget to 'flush', to get the answer.

In response, you will receive 1, if there exists a path going from (x_1, y_1) to (x_2, y_2) only to the right or down, numbers in cells of which form a palindrome, and 0 otherwise.

In case your query is invalid or you asked more than n^2 queries, program will print -1 and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

When you determine numbers in all cells, output "!".

Then output n lines, the i-th of which is a string of length n, corresponding to numbers in the i-th row of the grid.

After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

  * fflush(stdout) or cout.flush() in C++;
  * System.out.flush() in Java;
  * flush(output) in Pascal;
  * stdout.flush() in Python;
  * see documentation for other languages.



Hack Format

To hack, use the following format.

The first line should contain a single odd integer n (side of your grid).

The i-th of n following lines should contain a string of length n corresponding to the i-th row of the grid. Top left element of the grid has to be equal to 1, bottom right has to be equal to 0.

Example

Input


3
0
1
0
1
1
1
1

Output


? 1 1 1 3
? 1 1 2 3
? 2 1 2 3
? 3 1 3 3
? 2 2 3 3
? 1 2 3 2
? 1 2 3 3
!
100
001
000

Solution (in JAVA):

Solution:
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
What do you think? What do you think?
1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
*******************************
I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
300iq as writer = Sad!
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1205C
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int[][] grid = new int[N+1][N+1];
         for(int i=1; i <= N; i++)
            Arrays.fill(grid[i], -1);
         grid[1][1] = 1;
         grid[1][2] = grid[N][N] = 0;
         for(int i=3; i <= N; i++)
            grid[1][i] = query(1, i-2, 1, i, infile)^grid[1][i-2]^1;
         for(int r=2; r <= N; r++)
            for(int c=N; c >= 1; c--)
               if(grid[r][c] == -1)
               {
                  if(c >= 2)
                     grid[r][c] = query(r-1, c-1, r, c, infile)^grid[r-1][c-1]^1;
                  else
                     grid[r][c] = query(r, c, r, c+2, infile)^grid[r][c+2]^1;
               }
         ArrayList<Integer> vals = new ArrayList<Integer>();
         ArrayDeque<Integer> rq = new ArrayDeque<Integer>();
         ArrayDeque<Integer> cq = new ArrayDeque<Integer>();
         for(int r=1; r <= N; r++)
         {
            vals.add(grid[r][1]);
            rq.add(r);
            cq.add(1);
         }
         for(int c=2; c <= N; c++)
         {
            vals.add(grid[N][c]);
            rq.add(N);
            cq.add(c);
         }
         int flipped = 0;
         for(int head=0; head < vals.size()-3; head++)
         {
            if(good(vals, head))
            {
               int firstR = rq.poll();
               int firstC = cq.poll();
               rq.poll(); rq.poll();
               cq.poll(); cq.poll();
               int lastR = rq.poll();
               int lastC = cq.poll();
               int tag = query(firstR, firstC, lastR, lastC, infile);
               int palim = checkPalim(vals, head);
               flipped = tag^palim;
               break;
            }
            rq.poll();
            cq.poll();
         }
         for(int r=1; r <= N; r++)  
            for(int c=1; c <= N; c++)
               if((r+c)%2 == 1)
                  grid[r][c] ^= flipped;
         System.out.println("!");
         for(int r=1; r <= N; r++)
         {
            for(int c=1; c <= N; c++)
               System.out.print(grid[r][c]);
            System.out.println();
         }
      }
      public static int checkPalim(ArrayList<Integer> ls, int head)
      {
         if(ls.get(head) != ls.get(head+3) || ls.get(head+1) != ls.get(head+2))
            return 0;
         return 1;
      }
      public static boolean good(ArrayList<Integer> ls, int head)
      {
         int relationship1 = ls.get(head)^ls.get(head+2);
         int relationship2 = ls.get(head+1)^ls.get(head+3);
         return relationship1 == relationship2;
      }
      public static int query(int a, int b, int c, int d, BufferedReader infile) throws Exception
      {
         System.out.println("? "+a+" "+b+" "+c+" "+d);
         System.out.flush();
         return Integer.parseInt(infile.readLine());
      }
   }