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:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
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;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        FastScanner in;
        PrintWriter out;

        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            this.in = in;
            this.out = out;
            int n = in.nextInt();

            final int[] dr = {-2, -1, 0, 0, 1, 2};
            final int[] dc = {0, -1, -2, 2, 1, 0};
            int[] q = new int[n * n];
            q[0] = 0;
            q[1] = 1;
            int qt = 0;
            int qh = 2;
            boolean[] was = new boolean[n * n];
            int[] p = new int[n * n];
            was[q[0]] = true;
            was[q[1]] = true;
            p[q[0]] = -1;
            p[q[1]] = -1;
            p[n * n - 1] = -1;
            was[n * n - 1] = true;
            while (qt < qh) {
                int r = q[qt] / n;
                int c = q[qt] % n;
                ++qt;
                for (int dir = 0; dir < 6; dir++) {
                    int nr = r + dr[dir];
                    int nc = c + dc[dir];
                    if (nr < 0 || nr >= n || nc < 0 || nc >= n) {
                        continue;
                    }
                    int x = nr * n + nc;
                    if (!was[x]) {
                        was[x] = true;
                        q[qh++] = x;
                        p[x] = r * n + c;
                    }
                }
            }

            Graph g = new Graph(n * n);
            boolean[][] same = new boolean[g.n][g.n];
            addEdge(g, same, 0, n * n - 1, false);
            for (int i = 0; i < n * n; i++) {
                int j = p[i];
                if (j < 0) {
                    continue;
                }
                boolean s = send(n, i, j);
                addEdge(g, same, i, j, s);
            }

            int[] col = new int[n * n];
            Arrays.fill(col, -1);
            col[0] = 1;

            dfs(g, 0, col, same);
            col[1] = 0;
            dfs(g, 1, col, same);

            int[] f = findFour(n, col);
            if (f == null) {
                throw new AssertionError();
            }

            addEdge(g, same, f[0], f[1], send(n, f[0], f[1]));

            Arrays.fill(col, -1);
            col[0] = 1;
            dfs(g, 0, col, same);

            out.println("!");
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    out.print(col[n * i + j]);
                }
                out.println();
            }
            out.flush();
        }

        private boolean send(int n, int i, int j) {
            int r0 = i / n;
            int c0 = i % n;
            int r1 = j / n;
            int c1 = j % n;
            if (r0 + c0 < r1 + c1) {
                out.printf("? %d %d %d %d\n", r0 + 1, c0 + 1, r1 + 1, c1 + 1);
            } else {
                out.printf("? %d %d %d %d\n", r1 + 1, c1 + 1, r0 + 1, c0 + 1);
            }
            out.flush();
            return in.nextInt() == 1;
        }

        private int[] findFour(int n, int[] col) {
            List<Integer> path = new ArrayList<>();
            for (int r = 0; r < n; r++) {
                for (int c = 0; c < n; c++) {
                    int[] f = rec(n, col, path, r, c);
                    if (f != null) {
                        return f;
                    }
                }
            }
            return null;
        }

        private int[] rec(int n, int[] col, List<Integer> path, int r, int c) {
            path.add(r * n + c);
            if (path.size() == 4) {
                int xor = 0;
                for (int x : path) {
                    xor ^= col[x];
                }
                int[] f = xor == 0 ? new int[]{path.get(0), path.get(3)} : null;
                path.remove(path.size() - 1);
                return f;
            }
            if (r + 1 < n) {
                int[] f = rec(n, col, path, r + 1, c);
                if (f != null) {
                    return f;
                }
            }
            if (c + 1 < n) {
                int[] f = rec(n, col, path, r, c + 1);
                if (f != null) {
                    return f;
                }
            }
            path.remove(path.size() - 1);
            return null;
        }

        private void addEdge(Graph g, boolean[][] same, int i, int j, boolean s) {
            g.addEdge(i, j);
            g.addEdge(j, i);
            same[i][j] = s;
            same[j][i] = s;
        }

        private void dfs(Graph g, int v, int[] col, boolean[][] same) {
            for (int e = g.firstEdge[v]; e >= 0; e = g.edgeNxt[e]) {
                int u = g.edgeDst[e];
                if (col[u] >= 0) {
                    continue;
                }
                if (same[u][v]) {
                    col[u] = col[v];
                } else {
                    col[u] = col[v] ^ 1;
                }
                dfs(g, u, col, same);
            }
        }

        class Graph {
            int[] firstEdge;
            int[] edgeNxt;
            int[] edgeDst;
            int numEdges;
            int n;

            Graph(int n) {
                this.n = n;
                firstEdge = new int[n];
                Arrays.fill(firstEdge, -1);
                final int initialCapacity = (int) 1e6;
                edgeNxt = new int[initialCapacity];
                edgeDst = new int[initialCapacity];
                numEdges = 0;
            }

            public void addEdge(int a, int b) {
                int e = numEdges++;
                if (e >= edgeNxt.length) {
                    reallocate();
                }
                edgeNxt[e] = firstEdge[a];
                edgeDst[e] = b;
                firstEdge[a] = e;
            }

            private void reallocate() {
                int k = edgeNxt.length;
                k = 3 * k / 2;
                edgeNxt = Arrays.copyOf(edgeNxt, k);
                edgeDst = Arrays.copyOf(edgeDst, k);
            }

        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

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

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

    }
}