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:
/**
 * author: derrick20
 * created: 11/9/20 7:17 PM
 */
import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class E_PalindromicPaths {
    public static void main(String[] args) {
        sc = new FastScanner();
        out = new PrintWriter(System.out);
        
        int N = sc.nextInt();
        grid = new int[N][N];
        // get known vals
        for (int i = 0; i < N; i++) {
            Arrays.fill(grid[i], -1);
        }
        grid[0][0] = 1;
        grid[N - 1][N - 1] = 0;
        int determine = 0;
        for (int col = 0; col <= N - 3; col += 2) {
            relate(0, col, 0, col + 2);
            determine++;
        }
        for (int row = 0; row <= N - 3; row += 2) {
            relate(row, 0, row + 2, 0);
            determine++;
        }
        for (int row = 0; row <= N - 2; row++) {
            int start = row % 2;
            for (int col = start; col <= N - 3 + start; col += 2) {
                if (row == N - 2 && col == N - 2) continue;
                relate(row, col, row + 1, col + 1);
                determine++;
            }
        }
        int even = (N * N + 1) / 2;
        int odd = N * N - even;
        ASSERT(determine == even - 2);
        // get relationships
        // assign (0, 1) to be 0 arbitrarily
        grid[0][1] = 0;
        int relation = 0;
        for (int row = 0; row <= N - 1; row++) {
            int ct = relation;
            if (row % 2 == 0) {
                for (int col = 1; col <= N - 4; col += 2) {
                    relate(row, col, row, col + 2);
                    relation++;
                }
                if (row != N - 1) {
                    relate(row, N - 2, row + 1, N - 1);
                    relation++;
                }
            } else {
                for (int col = N - 1; col >= 2; col -= 2) {
                    relate(row, col, row, col - 2);
                    relation++;
                }
                relate(row, 0, row + 1, 1);
                relation++;
            }
//            System.out.println("Did " + (relation - ct) + " on " + row);
        }

        ASSERT(relation == odd - 1);
        // now, we need to find a position on diagonal such that it's
        // a 3x3 of 1 in top left, 0 in bottom right
        // this works by continuity, woah!
        int corner = 0;
        for (; corner <= N - 3; corner += 2) {
            if (grid[corner + 2][corner + 2] == 0) {
                break;
            }
        }
        int x = grid[corner][corner + 2];
        int y = grid[corner + 1][corner + 1];
        int c = grid[corner][corner + 1];
        int d = grid[corner + 1][corner + 2];
        boolean pal1 = query(corner, corner, corner + 1, corner + 2);
        boolean pal2 = query(corner, corner + 1, corner + 2, corner + 2);
        boolean flip;
        if (!pal1 && !pal2) {
            if (x == y) {
                if (x == 1) {
                    flip = d != 0;
                } else {
                    flip = c != 1;
                }
            } else {
                ASSERT(c != d);
                flip = d != 0;
            }
        } else if (pal1) {
            // if it isn't rightfully a 1, then we gotta flip
            flip = d != 1;
        } else {
            flip = c != 0;
        }
        out.println("!");
        for (int row = 0; row < N; row++) {
            StringBuilder line = new StringBuilder();
            for (int col = 0; col < N; col++) {
                if (flip && (row + col) % 2 == 1) {
                    grid[row][col] ^= 1;
                }
                line.append(grid[row][col]);
            }
            out.println(line);
        }
        out.close();
    }

    static FastScanner sc;
    static PrintWriter out;
    static int[][] grid;

    static void relate(int r1, int c1, int r2, int c2) {
        boolean pal;
        if (r1 > r2 || c1 > c2) {
            pal = query(r2, c2, r1, c1);
        } else {
            pal = query(r1, c1, r2, c2);
        }
        grid[r2][c2] = pal ? grid[r1][c1] : grid[r1][c1] ^ 1;
    }

    static boolean query(int r1, int c1, int r2, int c2) {
        out.printf("? %d %d %d %d\n", r1 + 1, c1 + 1, r2 + 1, c2 + 1);
        out.flush();
        int res = sc.nextInt();
        if (res == -1)
            ASSERT(false);
        return res == 1;
    }

    static class FastScanner {
        private int BS = 1 << 16;
        private char NC = (char) 0;
        private byte[] buf = new byte[BS];
        private int bId = 0, size = 0;
        private char c = NC;
        private double cnt = 1;
        private BufferedInputStream in;

        public FastScanner() {
            in = new BufferedInputStream(System.in, BS);
        }

        public FastScanner(String s) {
            try {
                in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
            } catch (Exception e) {
                in = new BufferedInputStream(System.in, BS);
            }
        }

        private char getChar() {
            while (bId == size) {
                try {
                    size = in.read(buf);
                } catch (Exception e) {
                    return NC;
                }
                if (size == -1) return NC;
                bId = 0;
            }
            return (char) buf[bId++];
        }

        public int nextInt() {
            return (int) nextLong();
        }

        public int[] nextInts(int N) {
            int[] res = new int[N];
            for (int i = 0; i < N; i++) {
                res[i] = (int) nextLong();
            }
            return res;
        }

        public long[] nextLongs(int N) {
            long[] res = new long[N];
            for (int i = 0; i < N; i++) {
                res[i] = nextLong();
            }
            return res;
        }

        public long nextLong() {
            cnt = 1;
            boolean neg = false;
            if (c == NC) c = getChar();
            for (; (c < '0' || c > '9'); c = getChar()) {
                if (c == '-') neg = true;
            }
            long res = 0;
            for (; c >= '0' && c <= '9'; c = getChar()) {
                res = (res << 3) + (res << 1) + c - '0';
                cnt *= 10;
            }
            return neg ? -res : res;
        }

        public double nextDouble() {
            double cur = nextLong();
            return c != '.' ? cur : cur + nextLong() / cnt;
        }

        public double[] nextDoubles(int N) {
            double[] res = new double[N];
            for (int i = 0; i < N; i++) {
                res[i] = nextDouble();
            }
            return res;
        }

        public String next() {
            StringBuilder res = new StringBuilder();
            while (c <= 32) c = getChar();
            while (c > 32) {
                res.append(c);
                c = getChar();
            }
            return res.toString();
        }

        public String nextLine() {
            StringBuilder res = new StringBuilder();
            while (c <= 32) c = getChar();
            while (c != '\n') {
                res.append(c);
                c = getChar();
            }
            return res.toString();
        }

        public boolean hasNext() {
            if (c > 32) return true;
            while (true) {
                c = getChar();
                if (c == NC) return false;
                else if (c > 32) return true;
            }
        }
    }
    static void ASSERT(boolean assertion, String message) {
        if (!assertion) throw new AssertionError(message);
    }
    static void ASSERT(boolean assertion) {
        if (!assertion) throw new AssertionError();
    }
}