The only difference between the easy and hard versions is that tokens of type O do not appear in the input of the easy version.

Errichto gave Monogon the following challenge in order to intimidate him from taking his top contributor spot on Codeforces.

In a Tic-Tac-Toe grid, there are n rows and n columns. Each cell of the grid is either empty or contains a token. There are two types of tokens: X and O. If there exist three tokens of the same type consecutive in a row or column, it is a winning configuration. Otherwise, it is a draw configuration.

<image> The patterns in the first row are winning configurations. The patterns in the second row are draw configurations. 

In an operation, you can change an X to an O, or an O to an X. Let k denote the total number of tokens in the grid. Your task is to make the grid a draw in at most ⌊ k/3⌋ (rounding down) operations.

You are not required to minimize the number of operations.

Input

The first line contains a single integer t (1≤ t≤ 100) — the number of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 300) — the size of the grid.

The following n lines each contain a string of n characters, denoting the initial grid. The character in the i-th row and j-th column is '.' if the cell is empty, or it is the type of token in the cell: 'X' or 'O'.

It is guaranteed that not all cells are empty.

In the easy version, the character 'O' does not appear in the input.

The sum of n across all test cases does not exceed 300.

Output

For each test case, print the state of the grid after applying the operations.

We have proof that a solution always exists. If there are multiple solutions, print any.

Example

Input


3
3
.X.
XXX
.X.
6
XX.XXX
XXXXXX
XXX.XX
XXXXXX
XX.X.X
XXXXXX
5
XXX.X
.X..X
XXX.X
..X..
..X..


Output


.X.
XOX
.X.
XX.XXO
XOXXOX
OXX.XX
XOOXXO
XX.X.X
OXXOXX
XOX.X
.X..X
XXO.O
..X..
..X..

Note

In the first test case, there are initially three 'X' consecutive in the second row and the second column. By changing the middle token to 'O' we make the grid a draw, and we only changed 1≤ ⌊ 5/3⌋ token.

In the second test case, we change only 9≤ ⌊ 32/3⌋ tokens, and there does not exist any three 'X' or 'O' consecutive in a row or column, so it is a draw.

In the third test case, we change only 3≤ ⌊ 12/3⌋ tokens, and the resulting grid is a draw.
import java.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class CF1450C {
    public static void main(String[] args) throws IOException {
        int t = ri();
        next: while (t --> 0) {
            int n = ri(), cnt = 0, X = 0;
            char a[][] = new char[n][n];
            for (int i = 0; i < n; ++i) {
                a[i] = rcha();
            }
            for (char[] b : a) {
                for (char c : b) {
                    if (c == 'X') ++X;
                }
            }

            for (int k = 0; k <= 2; ++k) {
                cnt = 0;
                char b[][] = new char[n][n];
                for (int i = 0; i < n; ++i) {
                    for (int j = 0; j < n; ++j) {
                        b[i][j] = a[i][j];
                    }
                }
                for (int i = 0; i < n; ++i) {
                    for (int j = (i + k) % 3; j < n; j += 3) {
                        if (b[i][j] == 'X') {
                            b[i][j] = 'O';
                            ++cnt;
                        }
                    }
                }
                if (cnt <= X / 3) {
                    for (int i = 0; i < n; ++i) {
                        prln(b[i]);
                    }
                    continue next;
                }
            }
        }
        close();
    }

    static BufferedReader __in = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter __out = new PrintWriter(new OutputStreamWriter(System.out));
    static StringTokenizer input;

    // references
    // IBIG = 1e9 + 7
    // IRAND ~= 3e8
    // IMAX ~= 2e10
    // LMAX ~= 9e18
    // constants
    static final int IBIG = 1000000007;
    static final int IRAND = 327859546;
    static final int IMAX = 2147483647;
    static final int IMIN = -2147483648;
    static final long LMAX = 9223372036854775807L;
    static final long LMIN = -9223372036854775808L;
    // util
    static int minof(int a, int b, int c) {return min(a, min(b, c));}
    static int minof(int... x) {return x.length == 1 ? x[0] : x.length == 2 ? min(x[0], x[1]) : min(x[0], minstarting(1, x));}
    static int minstarting(int offset, int... x) {assert x.length > 2; return offset == x.length - 2 ? min(x[offset], x[offset + 1]) : min(x[offset], minstarting(offset + 1, x));}
    static long minof(long a, long b, long c) {return min(a, min(b, c));}
    static long minof(long... x) {return x.length == 1 ? x[0] : x.length == 2 ? min(x[0], x[1]) : min(x[0], minstarting(1, x));}
    static long minstarting(int offset, long... x) {assert x.length > 2; return offset == x.length - 2 ? min(x[offset], x[offset + 1]) : min(x[offset], minstarting(offset + 1, x));}
    static int maxof(int a, int b, int c) {return max(a, max(b, c));}
    static int maxof(int... x) {return x.length == 1 ? x[0] : x.length == 2 ? max(x[0], x[1]) : max(x[0], maxstarting(1, x));}
    static int maxstarting(int offset, int... x) {assert x.length > 2; return offset == x.length - 2 ? max(x[offset], x[offset + 1]) : max(x[offset], maxstarting(offset + 1, x));}
    static long maxof(long a, long b, long c) {return max(a, max(b, c));}
    static long maxof(long... x) {return x.length == 1 ? x[0] : x.length == 2 ? max(x[0], x[1]) : max(x[0], maxstarting(1, x));}
    static long maxstarting(int offset, long... x) {assert x.length > 2; return offset == x.length - 2 ? max(x[offset], x[offset + 1]) : max(x[offset], maxstarting(offset + 1, x));}
    static int powi(int a, int b) {if(a == 0) return 0; int ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static long powl(long a, int b) {if(a == 0) return 0; long ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static int floori(double d) {return (int)d;}
    static int ceili(double d) {return (int)ceil(d);}
    static long floorl(double d) {return (long)d;}
    static long ceill(double d) {return (long)ceil(d);}
    // input
    static void r() throws IOException {input = new StringTokenizer(__in.readLine());}
    static int ri() throws IOException {return Integer.parseInt(__in.readLine());}
    static long rl() throws IOException {return Long.parseLong(__in.readLine());}
    static int[] ria(int n) throws IOException {int[] a = new int[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Integer.parseInt(input.nextToken()); return a;}
    static long[] rla(int n) throws IOException {long[] a = new long[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Long.parseLong(input.nextToken()); return a;}
    static char[] rcha() throws IOException {return __in.readLine().toCharArray();}
    static String rline() throws IOException {return __in.readLine();}
    static int rni() throws IOException {input = new StringTokenizer(__in.readLine()); return Integer.parseInt(input.nextToken());}
    static int ni() {return Integer.parseInt(input.nextToken());}
    static long rnl() throws IOException {input = new StringTokenizer(__in.readLine()); return Long.parseLong(input.nextToken());}
    static long nl() {return Long.parseLong(input.nextToken());}
    // output
    static void pr(int i) {__out.print(i);}
    static void prln(int i) {__out.println(i);}
    static void pr(long l) {__out.print(l);}
    static void prln(long l) {__out.println(l);}
    static void pr(double d) {__out.print(d);}
    static void prln(double d) {__out.println(d);}
    static void pr(char c) {__out.print(c);}
    static void prln(char c) {__out.println(c);}
    static void pr(char[] s) {__out.print(new String(s));}
    static void prln(char[] s) {__out.println(new String(s));}
    static void pr(String s) {__out.print(s);}
    static void prln(String s) {__out.println(s);}
    static void pr(Object o) {__out.print(o);}
    static void prln(Object o) {__out.println(o);}
    static void prln() {__out.println();}
    static void pryes() {__out.println("yes");}
    static void pry() {__out.println("Yes");}
    static void prY() {__out.println("YES");}
    static void prno() {__out.println("no");}
    static void prn() {__out.println("No");}
    static void prN() {__out.println("NO");}
    static void pryesno(boolean b) {__out.println(b ? "yes" : "no");};
    static void pryn(boolean b) {__out.println(b ? "Yes" : "No");}
    static void prYN(boolean b) {__out.println(b ? "YES" : "NO");}
    static void prln(int... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
    static void prln(long... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
    static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for(int i = 0; i < n; __out.print(iter.next()), __out.print(' '), ++i); __out.println(iter.next());}
    static void flush() {__out.flush();}
    static void close() {__out.close();}}