#include <bits/stdc++.h>
using namespace std;
#define FAST ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define int               long long
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define rall(x)           x.rbegin(), x.rend()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define repi(i,a,b)       for(int i=b-1;i>=a;i--)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               __builtin_popcount
#define ppcll             __builtin_popcountll
#define sz5               100005
#define sz6               1000005
#define V(a)              vector<a>
#define MP               make_pair
#define MT               make_tuple
#define G(a,b)           get<a>(b)
const long long INF = 1e18;
const int32_t M = 1e9  + 7 ;
const int32_t MM = 998244353;
const int N = 5e5 + 5;
template<typename T1, typename T2>istream& operator>>(istream& in, pair<T1, T2> &a) {in >> a.fr >> a.sc; return in;}
template<typename T1, typename T2>ostream& operator<<(ostream& out, pair<T1, T2> a) {out << a.fr << " " << a.sc; return out;}
template<typename T, typename T1>T amax(T &a, T1 b) {if (b > a)a = b; return a;}
template<typename T, typename T1>T amin(T &a, T1 b) {if (b < a)a = b; return a;}

void solve()
{
	int n;
	cin >> n;
	char arr[n][n];
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			cin >> arr[i][j];
		}
	}
	// for (int i = 0; i < n; i++)
	// {
	// 	for (int j = 0; j < n; j++)
	// 	{
	// 		cout << arr[i][j] << " ";
	// 	}
	// 	cout << endl;
	// }

	pair<int, int> a;
	a.first = -1;
	a.second = -1;
	pair<int, int> b;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (arr[i][j] == '*')
			{
				if (a.first == -1)
				{
					a.first = i;
					a.second = j;
				}
				else
				{
					b.first = i;
					b.second = j;
					break;
				}
			}
		}
	}

	if (a.first == b.first)
	{
		if (a.first != 0)
		{
			arr[0][a.second]  = '*';
			arr[0][b.second] = '*';
		}
		else
		{
			arr[1][a.second]  = '*';
			arr[1][b.second] = '*';
		}
	}
	else if (a.second == b.second)
	{
		if (a.second != 0)
		{
			arr[a.first][0]  = '*';
			arr[b.first][0] = '*';
		}
		else
		{
			arr[a.first][1]  = '*';
			arr[b.first][1] = '*';
		}
	}
	else
	{
		arr[a.first][b.second] = '*';
		arr[b.first][a.second] = '*';
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			cout << arr[i][j];
		}
		cout << endl;
	}



}
signed main()
{
#ifndef ONLINE_JUDGE
	//FOR GETTING INPUT FROM INPUT.TEXT
	freopen("input", "r", stdin);
	//for getting output to output.txt
	freopen("output", "w", stdout);
#endif
	FAST
	int t;
	cin >> t;
	while (t--)
	{
		solve();
	}

	return 0;
}
### Prompt

Please formulate a cpp solution to the following problem:

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


*.*.
....
*.*.
....
**
**
**
**
*.*
*.*
...
.....
.**..
.....
.**..
.....
....
....
**..
**..

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define FAST ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define int               long long
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define rall(x)           x.rbegin(), x.rend()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define repi(i,a,b)       for(int i=b-1;i>=a;i--)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               __builtin_popcount
#define ppcll             __builtin_popcountll
#define sz5               100005
#define sz6               1000005
#define V(a)              vector<a>
#define MP               make_pair
#define MT               make_tuple
#define G(a,b)           get<a>(b)
const long long INF = 1e18;
const int32_t M = 1e9  + 7 ;
const int32_t MM = 998244353;
const int N = 5e5 + 5;
template<typename T1, typename T2>istream& operator>>(istream& in, pair<T1, T2> &a) {in >> a.fr >> a.sc; return in;}
template<typename T1, typename T2>ostream& operator<<(ostream& out, pair<T1, T2> a) {out << a.fr << " " << a.sc; return out;}
template<typename T, typename T1>T amax(T &a, T1 b) {if (b > a)a = b; return a;}
template<typename T, typename T1>T amin(T &a, T1 b) {if (b < a)a = b; return a;}

void solve()
{
	int n;
	cin >> n;
	char arr[n][n];
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			cin >> arr[i][j];
		}
	}
	// for (int i = 0; i < n; i++)
	// {
	// 	for (int j = 0; j < n; j++)
	// 	{
	// 		cout << arr[i][j] << " ";
	// 	}
	// 	cout << endl;
	// }

	pair<int, int> a;
	a.first = -1;
	a.second = -1;
	pair<int, int> b;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (arr[i][j] == '*')
			{
				if (a.first == -1)
				{
					a.first = i;
					a.second = j;
				}
				else
				{
					b.first = i;
					b.second = j;
					break;
				}
			}
		}
	}

	if (a.first == b.first)
	{
		if (a.first != 0)
		{
			arr[0][a.second]  = '*';
			arr[0][b.second] = '*';
		}
		else
		{
			arr[1][a.second]  = '*';
			arr[1][b.second] = '*';
		}
	}
	else if (a.second == b.second)
	{
		if (a.second != 0)
		{
			arr[a.first][0]  = '*';
			arr[b.first][0] = '*';
		}
		else
		{
			arr[a.first][1]  = '*';
			arr[b.first][1] = '*';
		}
	}
	else
	{
		arr[a.first][b.second] = '*';
		arr[b.first][a.second] = '*';
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			cout << arr[i][j];
		}
		cout << endl;
	}



}
signed main()
{
#ifndef ONLINE_JUDGE
	//FOR GETTING INPUT FROM INPUT.TEXT
	freopen("input", "r", stdin);
	//for getting output to output.txt
	freopen("output", "w", stdout);
#endif
	FAST
	int t;
	cin >> t;
	while (t--)
	{
		solve();
	}

	return 0;
}
```