There is a square field of size n × n in which two cells are marked. These cells can be in the same row or column.

You are to mark two more cells so that they are the corners of a rectangle with sides parallel to the coordinate axes.

For example, if n=4 and a rectangular field looks like this (there are asterisks in the marked cells):

$$$ \begin{matrix} . & . & * & . \\\ . & . & . & . \\\ * & . & . & . \\\ . & . & . & . \\\ \end{matrix} $$$

Then you can mark two more cells as follows

$$$ \begin{matrix} * & . & * & . \\\ . & . & . & . \\\ * & . & * & . \\\ . & . & . & . \\\ \end{matrix} $$$

If there are several possible solutions, then print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 400). Then t test cases follow.

The first row of each test case contains a single integer n (2 ≤ n ≤ 400) — the number of rows and columns in the table.

The following n lines each contain n characters '.' or '*' denoting empty and marked cells, respectively.

It is guaranteed that the sums of n for all test cases do not exceed 400.

It is guaranteed that there are exactly two asterisks on the field. They can be in the same row/column.

It is guaranteed that the solution exists.

Output

For each test case, output n rows of n characters — a field with four asterisks marked corresponding to the statements. If there multiple correct answers, print any of them.

Example

Input


6
4
..*.
....
*...
....
2
*.
.*
2
.*
.*
3
*.*
...
...
5
.....
..*..
.....
.*...
.....
4
....
....
*...
*...


Output


*.*.
....
*.*.
....
**
**
**
**
*.*
*.*
...
.....
.**..
.....
.**..
.....
....
....
**..
**..
#include <bits/stdc++.h>
using namespace std;
void solve()
{
	int n;cin >> n;
	vector<string> input(n);
	pair<int, int> p1 = {-1, -1}, p2 = {-1, -1};
	for (int i = 0 ;i < n ;i++)
	{
		cin >> input[i];
		for (int j = 0 ; j < n ; j++)
		{
			string row = input[i];
			if ( row[j] == '*' )
			{
				if (p1.first == -1) p1 = {i, j};
				else p2 = {i, j};
			}
		}
	}

	if (p1.first != p2.first && p1.second != p2.second)
	{
		input[p2.first][p1.second] = '*';
		input[p1.first][p2.second] = '*';
	}else if (p1.first != p2.first && p1.second == p2.second)
	{
		input[p1.first][(p1.second + 1) % n] = '*';
		input[p2.first][(p1.second + 1) % n] = '*';
	}else
	{
		input[(p1.first + 1) % n][p1.second] = '*';
		input[(p1.first + 1) % n][p2.second] = '*';
	}
	for (int i = 0 ; i < n ; i++) cout << input[i] << endl;

}
int main(int argc, char const *argv[])
{
	int t;cin >> t;
	while(t--) solve();
	return 0;
}