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).

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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.IOException;
import java.io.InputStream;
import java.io.OutputStream;

public class Main {
    private final FastRead reader;
    private final FastWrite writer;

    private boolean ask(int x0, int y0, int x1, int y1) throws IOException {
        writer.println("? "  + (x0+1) + ' ' + (y0+1)
                       + ' ' + (x1+1) + ' ' + (y1+1));
        writer.flush();
        int res = reader.nextInt();
        if (res == -1) System.exit(0);
        return res == 1;
    }

    private boolean needRev(int[][] res) throws IOException {
        int bg = 0;
        while (res[bg][bg] == res[bg+2][bg+2]) ++bg;
        int[][] st = new int[][] {
                {res[bg  ][bg], res[bg  ][bg+1], res[bg  ][bg+2]},
                {res[bg+1][bg], res[bg+1][bg+1], res[bg+1][bg+2]},
                {res[bg+2][bg], res[bg+2][bg+1], res[bg+2][bg+2]},
        };

        if (st[0][1] != st[1][0]) {
            boolean eq = ask(bg, bg, bg+1, bg+2);
            return eq != (st[0][0] == st[1][2]);
        }

        if (st[1][2] != st[2][1]) {
            boolean eq = ask(bg, bg+1, bg+2, bg+2);
            return eq != (st[0][1] == st[2][2]);
        }

        if ((st[0][1] == st[1][2]) == (st[0][0] == st[1][1])) {
            boolean eq = ask(bg, bg, bg+1, bg+2);
            return eq != (st[0][0] == st[1][2]);
        } else {
            boolean eq = ask(bg, bg+1, bg+2, bg+2);
            return eq != (st[0][1] == st[2][2]);
        }
    }

    private void run() throws Exception {
        int n = reader.nextInt();
        int[][] res = new int[n][n];
        res[0][0] = 1;
        res[n-1][n-1] = 0;

        for (int i = 0; i < n; ++i) {
            for (int j = i&1; j < n; j += 2) {
                if (i == 0 && j == 0) continue;
                if (i+1 == n && j+1 == n) continue;
                int i0, j0;
                if (i >= 2) {
                    i0 = i-2; j0 = j;
                } else if (j >= 2) {
                    i0 = i; j0 = j-2;
                } else {
                    i0 = i-1; j0 = j-1;
                }
                int c0 = res[i0][j0];
                res[i][j] = ask(i0, j0, i, j) ? c0 : 1 - c0;
            }
        }

        /* init mask */ {
            /* init dif = -(n-2) */ {
                res[0][n-2] = 0;
                boolean eq = ask(0, n-2, 1, n-1);
                res[1][n-1] = eq ? 0 : 1;
            }
            for (int dif = -(n-4); dif <= n-2; dif += 2) {
                // i - j = dif - 1
                // i + j =   n - 1
                // i = (n - 1 + dif - 1) / 2 = (n+dif)/2 - 1
                // j = (n - 1 - dif + 1) / 2 = (n-dif)/2
                int mid_i = (n+dif)/2 - 1;
                int mid_j = (n-dif)/2;
                res[mid_i+1][mid_j] =
                        ask(mid_i-1, mid_j, mid_i+1, mid_j)
                        ? res[mid_i-1][mid_j] : 1 - res[mid_i-1][mid_j];
                int mxDown = Integer.min(mid_i+1, mid_j);
                for (int i = 1; i <= mxDown; ++i) {
                    int x = mid_i+1 - i;
                    int y = mid_j - i;
                    boolean eq = ask(x, y, x+1, y+1);
                    res[x][y] = eq ? res[x+1][y+1] : 1 - res[x+1][y+1];
                }
                int mxUp = mxDown - 1;
                for (int i = 1; i <= mxUp; ++i) {
                    int x = mid_i+1 + i;
                    int y = mid_j + i;
                    boolean eq = ask(x-1, y-1, x, y);
                    res[x][y] = eq ? res[x-1][y-1] : 1 - res[x-1][y-1];
                }
            }
        }

        if (needRev(res)) {
            for (int i = 0; i < n; ++i) {
                for (int j = 1 - i&1; j < n; j += 2) {
                    res[i][j] = 1 - res[i][j];
                }
            }
        }

        writer.println("!");
        for (int[] line : res) {
            for (int x : line)
                writer.print("" + x);
            writer.println();
        }
    }

    /* ====================== Template ====================== */
    public static void main(String[] args) throws Exception {
        Main m = new Main();
        try { m.run(); } finally { m.writer.flush(); }
    }

    private Main() {
        reader = new FastRead(System.in);
        writer = new FastWrite(System.out);
    }

    static final int BUFFER_SIZE = 1<<16;
    static class FastRead {
        private final InputStream in;
        private final byte[] buffer;
        private int cur, ready;

        private void refillBuffer() throws IOException {
            ready = in.read(buffer, 0, buffer.length);
            cur = (ready == -1) ? -1 : 0;
        }

        public FastRead(InputStream in) {
            this.in = in;
            buffer = new byte[BUFFER_SIZE];
            cur = ready = 0;
        }

        private byte read() throws IOException {
            if (cur == ready) refillBuffer();
            return buffer[cur++];
        }

        public String nextString(int maxLength) throws IOException {
            byte[] res = new byte[maxLength];
            int len =  0;
            byte c = read();
            while (c == ' ' || c == '\n' || c=='\r')
                c = read();
            while (c != ' ' && c != '\n' && c!='\r') {
                res[len++] = c;
                c = read();
            }
            return new String(res, 0, len);
        }

        public String nextString() throws IOException {
            byte[] res = new byte[64];
            int len =  0;
            byte c = read();
            while (c == ' ' || c == '\n' || c=='\r')
                c = read();
            while (c != ' ' && c != '\n' && c!='\r') {
                if (len == res.length) {
                    byte[] oldRes = res;
                    res = new byte[res.length * 4];
                    System.arraycopy(oldRes, 0, res, 0, oldRes.length);
                }
                res[len++] = c;
                c = read();
            }
            return new String(res, 0, len);
        }

        public int nextInt() throws IOException {
            int ret =  0;
            boolean neg;

            byte c = read();
            while (c <= ' ')
                c = read();
            neg = c == '-';
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
                c = read();
            } while (c > ' ');

            return neg ? -ret : ret;
        }

        public long nextLong() throws IOException {
            long ret =  0;
            boolean neg;

            byte c = read();
            while (c <= ' ')
                c = read();
            neg = c == '-';
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
                c = read();
            } while (c > ' ');

            return neg ? -ret : ret;
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(nextString(32));
        }

        public float nextFloat() throws IOException {
            return Float.parseFloat(nextString(32));
        }

        public void close() throws IOException {
            in.close();
        }
    }
    static class FastWrite {
        private final OutputStream out;
        private final byte[] buffer;
        private int ready;

        public FastWrite(OutputStream out) {
            this.out = out;
            buffer = new byte[BUFFER_SIZE];
            ready = 0;
        }

        public void flush() throws IOException {
            if (ready > 0) {
                out.write(buffer, 0, ready);
                ready = 0;
            }
        }

        private void put(byte b) throws IOException {
            if (ready == buffer.length) flush();
            buffer[ready++] = b;
        }

        private void put(byte[] b) throws IOException {
            int done = 0;
            if (ready < buffer.length) {
                done = buffer.length - ready;
                if (done > b.length) done = b.length;
                System.arraycopy(b, 0, buffer, ready, done);
                ready += done;
            }

            if (ready == buffer.length) flush();
            if (done == b.length) return;
            assert ready == 0;

            if ((b.length - done) * 2 < buffer.length) {
                ready = b.length - done;
                System.arraycopy(b, done, buffer, 0, ready);
            } else {
                out.write(b, done, b.length - done);
            }
        }

        public void print(String s) throws IOException {
            put(s.getBytes());
        }

        public void println(String s) throws IOException {
            put(s.getBytes()); put((byte)'\n');
        }

        public void println() throws IOException {
            put((byte)'\n');
        }
    }
}