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.
//Some of the methods are copied from GeeksforGeeks Website
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{ 
 public static void main (String[] args) throws java.lang.Exception
   {
	try{
	     //long n=sc.nextLong();
	     //String s=sc.next();
	     FastReader sc=new FastReader(System.in);
	     //Reader sc=new Reader();
	     PrintWriter out=new PrintWriter(System.out);
		  int t=sc.nextInt();
		  while(t-->0)
		    { int n=sc.nextInt();
		      char ar[][]=new char[n][n];
		      int rgb[]=new int[3];
		      for(int i=0;i<n;i++)
		       {
		           String s=sc.next();
		           for(int j=0;j<n;j++)
		             {
		                 ar[i][j]=s.charAt(j);
		                 if(ar[i][j]=='X')
		                 rgb[(i+j)%3]++;
		             }
		       }
		  //for(int e : rgb)
		  // out.print(e+" "); 
		  //out.println();
		    int index=0;
		    int min=Integer.MAX_VALUE;
		    for(int i=0;i<3;i++)
		      {
		          if(rgb[i]<min)
		           {
		               min=rgb[i];
		               index=i;
		           }
		      }
		      
		      for(int i=0;i<n;i++)
		        {
		            for(int j=0;j<n;j++)
		              {
		                if((i+j)%3==index && ar[i][j]=='X')
		                 ar[i][j]='O';
		              }
		        }
		     
		     
		     
		     for( int i=0;i<n;i++)
		        {
		            for(int j=0;j<n;j++)
		              {
		                out.print(ar[i][j]);  
		              }
		          out.println();
		        }
		    }
	     out.flush();
	     out.close();
	   }     
	catch(Exception e)
		 {}
	}
	
	



//Soluntion Ends Here........................................................................................


	//Template
	
   // int a[]=new int[n];
   // for(int i=0;i<n;i++)
   //  a[i]=sc.nextInt();
   
   // Map<Integer,Integer> map=new HashMap<>();
   // for(int i=0;i<n;i++)
   //  {
   //     if(!map.containsKey(a[i]))
   //       map.put(a[i],1);
   //     else
   //       map.replace(a[i],map.get(a[i])+1);
   //  }
     
  //ArrayList<Integer> al=new ArrayList<>();
  
 static int gcd(int a, int b) 
    { 
        if (a == 0) 
            return b; 
          
        return gcd(b%a, a); 
    } 
    
static void calcsubarray(long a[],long x[], int n, int c) 
 { 
    for (int i=0; i<(1<<n); i++) 
    { 
        long s = 0; 
        for (int j=0; j<n; j++) 
            if ((i & (1<<j))==0)
                s += a[j+c]; 
        x[i] = s; 
    } 
 } 
 
   static class FastReader{
 
        byte[] buf = new byte[2048];
        int index, total;
        InputStream in;
 
        FastReader(InputStream is) {
            in = is;
        }
 
        int scan() throws IOException {
            if (index >= total) {
                index = 0;
                total = in.read(buf);
                if (total <= 0) {
                    return -1;
                }
            }
            return buf[index++];
        }
 
        String next() throws IOException {
            int c;
            for (c = scan(); c <= 32; c = scan());
            StringBuilder sb = new StringBuilder();
            for (; c > 32; c = scan()) {
                sb.append((char) c);
            }
            return sb.toString();
        }
 
        int nextInt() throws IOException {
            int c, val = 0;
            for (c = scan(); c <= 32; c = scan());
            boolean neg = c == '-';
            if (c == '-' || c == '+') {
                c = scan();
            }
            for (; c >= '0' && c <= '9'; c = scan()) {
                val = (val << 3) + (val << 1) + (c & 15);
            }
            return neg ? -val : val;
        }
 
        long nextLong() throws IOException {
            int c;
            long val = 0;
            for (c = scan(); c <= 32; c = scan());
            boolean neg = c == '-';
            if (c == '-' || c == '+') {
                c = scan();
            }
            for (; c >= '0' && c <= '9'; c = scan()) {
                val = (val << 3) + (val << 1) + (c & 15);
            }
            return neg ? -val : val;
        }
    }
   
    static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 

}
// Thank You !