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.math.BigInteger;
import java.util.*;
import java.util.concurrent.TimeUnit;

public class c1450 implements Runnable{
	
    public static void main(String[] args) {
    	try{
            new Thread(null, new c1450(), "process", 1<<26).start();
        }
        catch(Exception e){
            System.out.println(e);
        }
    }
	public void run() {
		FastReader scan = new FastReader();
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		//PrintWriter out = new PrintWriter("file.out");
		Task solver = new Task();
		int t = scan.nextInt();
		//int t = 1;
		for(int i = 1; i <= t; i++) solver.solve(i, scan, out);
		out.close();
	}

	static class Task {
		static final int oo = Integer.MAX_VALUE;
		static final long OO = Long.MAX_VALUE;
		
		int N;
		public void solve(int testNumber, FastReader sc, PrintWriter out) {
			N = sc.nextInt();
			char[][] board = new char[N][N];
			for(int i = 0; i < N; i++) {
				board[i] = sc.nextLine().toCharArray();
			}
			
			int min = oo;
			int kVal = 0;
			for(int k = 0; k < 3; k++) {
				int count = 0;
				for(int i = 0; i < N; i++) {
					for(int j = (i+k) % 3; j < N; j+= 3) {
						if(board[i][j] == 'X') {
							count++;
						}
					}
				}
				if(count < min) {
					min = count;
					kVal = k;
				}
			}
			
			for(int i = 0; i < N; i++) {
				for(int j = (i+kVal) % 3; j < N; j+= 3) {
					if(board[i][j] == 'X') {
						board[i][j] = 'O';
					}
				}
			}
			
			for(char[] each: board) {
				for(char ea: each) {
					out.print(ea);
				}
				out.println();
			}
		}
		
	}
	static long modInverse(long N, long MOD) {
		return binpow(N, MOD - 2, MOD);
	}
	static long modDivide(long a, long b, long MOD) {
		a %= MOD;
		return (binpow(b, MOD-2, MOD) * a) % MOD;
	}
	static long binpow(long a, long b, long m) {
		a %= m;
		long res = 1;
		while (b > 0) {
			if ((b & 1) == 1)
				res = res * a % m;
			a = a * a % m;
			b >>= 1;
		}
		return res;
	}
	static int[] reverse(int a[]) 
    { 
        int[] b = new int[a.length]; 
        for (int i = 0, j = a.length; i < a.length; i++, j--) { 
            b[j - 1] = a[i]; 
        } 
        
        return b;
    }
	static long[] reverse(long a[]) 
    { 
        long[] b = new long[a.length]; 
        for (int i = 0, j = a.length; i < a.length; i++, j--) { 
            b[j - 1] = a[i]; 
        } 
        
        return b;
    }
	
	static void shuffle(Object[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			Object temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}
	
	static void shuffle(int[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			int temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}
	static void shuffle(long[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			long temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}
	
	static class tup implements Comparable<tup>, Comparator<tup>{
		int a, b;
		tup(int a,int b){
			this.a=a;
			this.b=b;
		}
		public tup() {
		}
		@Override
		public int compareTo(tup o){
			return Integer.compare(b,o.b);
		}
		@Override
		public int compare(tup o1, tup o2) {
			return Integer.compare(o1.b, o2.b);
		}
		
		@Override
	    public int hashCode() {
			return Objects.hash(a, b);
	    }
 
	    @Override
	    public boolean equals(Object obj) {
	    	if (this == obj)
                return true;
	    	if (obj == null)
                return false;
	    	if (getClass() != obj.getClass())
                return false;
	    	tup other = (tup) obj;
	    	return a==other.a && b==other.b;
	    }
	    
	    @Override
	    public String toString() {
	    	return a + " " + b;
	    }
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public FastReader(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(new File(s)));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
		
		int[] readArray(int size) {
			int[] a = new int[size];
			for(int i = 0; i < size; i++) {
				a[i] = nextInt();
			}
			return a;
		}
		
		long[] readLongArray(int size) {
			long[] a = new long[size];
			for(int i = 0; i < size; i++) {
				a[i] = nextLong();
			}
			return a;
		}
	}
	
	static void dbg(int[] arr) {
		System.out.println(Arrays.toString(arr));
	}
	static void dbg(long[] arr) {
		System.out.println(Arrays.toString(arr));
	}
	static void dbg(boolean[] arr) {
		System.out.println(Arrays.toString(arr));
	}

	static void dbg(Object... args) {
        for (Object arg : args)
            System.out.print(arg + " ");
        System.out.println();
    }
}