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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
 
public class C {
	
	final static int MAXN = 100_005;
	final static long MOD = (long) 1e9 + 7;
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		for(int tt =0 ;tt < t ; tt++) {
			int n = sc.nextInt();
			char matrix[][] = new char[n][n];
			char matrixs[][][] = new char[3][n][n];
			for(int i = 0 ; i < n ; i++) {
				String s = sc.next();
				matrix[i] = s.toCharArray();
				for(int j = 0 ; j < 3 ;j++) {
					matrixs[j][i] = s.toCharArray();
				}
			}
			
			
			int changes[] = new int[3];
			for(int x = 0 ; x < n ; x++) {
				for(int y = 0 ; y < n ; y++) {
					if(matrix[x][y] == '.')continue;
					changes[(x+y)%3]++;
					matrixs[(x+y)%3][x][y] = 'O';
				}
			}
			char ans[][];
			if (changes[0]<=changes[1] && changes[0] <= changes[2]) {
				ans=matrixs[0];
			}
			else if (changes[1]<=changes[2]) {
				ans=matrixs[1];
			}
			else {
				ans=matrixs[2];
			}
			StringBuilder sb = new StringBuilder();
			for(int i = 0 ; i < n ; i++) {
				for(int j = 0 ; j < n ; j++) {
					sb.append(ans[i][j]);
				}if(i != n-1) {
					sb.append("\n");
				}
			}
			System.out.println(sb);
		}
	}
	
	
	public static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}

    public static long[] factorial;
 
    public static void setFactorial() {
        factorial = new long[MAXN];
        factorial[0] = 1;
        for (int i = 1; i < MAXN; ++i) factorial[i] = factorial[i - 1] * i % MOD;
    }
 
    public static long getFactorial(int n) {
        if (factorial == null) setFactorial();
        return factorial[n];
    }
 
    public static long ncr(int n, int r) {
        if (r > n) return 0;
        if (factorial == null) setFactorial();
        long numerator = factorial[n];
        long denominator = factorial[r] * factorial[n - r] % MOD;
        return numerator * pow(denominator, MOD - 2, MOD) % MOD;
    }
 
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }
 
    public static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
    public static long pow(long base, long exp, long MOD) {
        base %= MOD;
        long ret = 1;
        while (exp > 0) {
            if ((exp & 1) == 1) ret = ret * base % MOD;
            base = base * base % MOD;
            exp >>= 1;
        }
        return ret;
    }
 
 
    public static long max(long... ar) {
        long ret = ar[0];
        for (long itr : ar) ret = Math.max(ret, itr);
        return ret;
    }
 
    public static int max(int... ar) {
        int ret = ar[0];
        for (int itr : ar) ret = Math.max(ret, itr);
        return ret;
    }
 
    public static long min(long... ar) {
        long ret = ar[0];
        for (long itr : ar) ret = Math.min(ret, itr);
        return ret;
    }
 
    public static int min(int... ar) {
        int ret = ar[0];
        for (int itr : ar) ret = Math.min(ret, itr);
        return ret;
    }
 
 
    public static long sum(long... ar) {
        long sum = 0;
        for (long itr : ar) sum += itr;
        return sum;
    }
 
    public static long sum(int... ar) {
        long sum = 0;
        for (int itr : ar) sum += itr;
        return sum;
    }
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		
		public String next() {
			while (!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());
		}
		
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
	}
 
}


