Problem: There are many sunflowers in the Garden of the Sun.

Garden of the Sun is a rectangular table with n rows and m columns, where the cells of the table are farmlands. All of the cells grow a sunflower on it. Unfortunately, one night, the lightning stroke some (possibly zero) cells, and sunflowers on those cells were burned into ashes. In other words, those cells struck by the lightning became empty. Magically, any two empty cells have no common points (neither edges nor corners).

Now the owner wants to remove some (possibly zero) sunflowers to reach the following two goals: 

  * When you are on an empty cell, you can walk to any other empty cell. In other words, those empty cells are connected. 
  * There is exactly one simple path between any two empty cells. In other words, there is no cycle among the empty cells. 



You can walk from an empty cell to another if they share a common edge.

Could you please give the owner a solution that meets all her requirements?

Note that you are not allowed to plant sunflowers. You don't need to minimize the number of sunflowers you remove. It can be shown that the answer always exists.

Input

The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases. The description of the test cases follows.

The first line contains two integers n, m (1 ≤ n,m ≤ 500) — the number of rows and columns. 

Each of the next n lines contains m characters. Each character is either 'X' or '.', representing an empty cell and a cell that grows a sunflower, respectively.

It is guaranteed that the sum of n ⋅ m for all test cases does not exceed 250 000.

Output

For each test case, print n lines. Each should contain m characters, representing one row of the table. Each character should be either 'X' or '.', representing an empty cell and a cell with a sunflower, respectively.

If there are multiple answers, you can print any. It can be shown that the answer always exists.

Example

Input


5
3 3
X.X
...
X.X
4 4
....
.X.X
....
.X.X
5 5
.X...
....X
.X...
.....
X.X.X
1 10
....X.X.X.
2 2
..
..


Output


XXX
..X
XXX
XXXX
.X.X
.X..
.XXX
.X...
.XXXX
.X...
.X...
XXXXX
XXXXXXXXXX
..
..

Note

Let's use (x,y) to describe the cell on x-th row and y-th column.

In the following pictures white, yellow, and blue cells stand for the cells that grow a sunflower, the cells lightning stroke, and the cells sunflower on which are removed, respectively.

<image>

In the first test case, one possible solution is to remove sunflowers on (1,2), (2,3) and (3 ,2). 

<image>

Another acceptable solution is to remove sunflowers on (1,2), (2,2) and (3,2). 

<image>

This output is considered wrong because there are 2 simple paths between any pair of cells (there is a cycle). For example, there are 2 simple paths between (1,1) and (3,3).

  1. (1,1)→ (1,2)→ (1,3)→ (2,3)→ (3,3)

  2. (1,1)→ (2,1)→ (3,1)→ (3,2)→ (3,3) 



<image>

This output is considered wrong because you can't walk from (1,1) to (3,3).

Solution (in CPP):

Solution:
//Original Code:
//#include <self/utility>
//#include <self/debug>
//using namespace std;
//int n,m;
//char c[505][505];
//
//void solve()
//{
//	scanf("%d%d",&n,&m);
//	for(int i=0;i<n;i++)
//	{
//		scanf("%s",c[i]);
//	}
//	for(int i=0;i<n;i+=3)
//	{
//		for(int j=0;j<m;j++) c[i][j]='X';
//	}
//	for(int i=3;i<n;i+=3)
//	{
//		int flag=false;
//		for(int z=i-2;z<i;z++)
//		{
//			for(int j=0;j<m;j++)
//			{
//				if(c[z][j]=='X' && !flag)
//				{
//					c[i-1+i-2-z][j]='X';
//					flag=true;
//				}
//			}
//		}
//		if(!flag)
//		{
//			c[i-2][0]=c[i-1][0]='X';
//		}
//	}
//	if(n%3==0)
//	{
//		for(int j=0;j<m;j++)
//		{
//			if(c[n-2][j]=='X' || c[n-1][j]=='X')
//			{
//				c[n-2][j]='X';
//				c[n-1][j]='X';
//			}
//		}
//	}
//	for(int i=0;i<n;i++) printf("%s\n",c[i]);
//}
//
//int main()
//{
//	// freopen("input.txt","r",stdin);
//	int T;
//	scanf("%d",&T);
//	while(T--) solve();
//	return 0;
//}

//substituted with C++ Inliner
#ifndef _SELF_UTILITY
#define _SELF_UTILITY 1
#include <numeric>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <string.h>
#include <stack>
#include <assert.h>
#include <bitset>
#include <time.h>
#define Endl endl
#define mp make_pair
#define mt make_tuple
#define ll long long 
#define ull unsigned long long
#define pii pair<int,int>
#define over(A) {cout<<A<<endl;exit(0);}
#define all(A) A.begin(),A.end()
#define quickcin ios_base::sync_with_stdio(false);
#ifdef __TAKE_MOD
int mod;
#else
#ifdef __TAKE_CONST_MOD
const int mod=__TAKE_CONST_MOD;
#else
const int mod=1000000007;
#endif
#endif
const int gmod=3;
const int inf=1039074182;
const double eps=1e-9;
const ll llinf=2LL*inf*inf;
template <typename T1,typename T2> inline void chmin(T1 &x,T2 b) {if(b<x) x=b;}
template <typename T1,typename T2> inline void chmax(T1 &x,T2 b) {if(b>x) x=b;}
inline void chadd(int &x,int b) {x+=b-mod;x+=(x>>31 & mod);}
template <typename T1,typename T2> inline void chadd(T1 &x,T2 b) {x+=b;if(x>=mod) x-=mod;}
template <typename T1,typename T2> inline void chmul(T1 &x,T2 b) {x=1LL*x*b%mod;}
template <typename T1,typename T2> inline void chmod(T1 &x,T2 b) {x%=b,x+=b;if(x>=b) x-=b;}
template <typename T> inline T mabs(T x) {return (x<0?-x:x);}
using namespace std;
#endif
#ifndef _SELF_DEBUG
#define _SELF_DEBUG 1
#ifndef _SELF_OPERATOR
#define _SELF_OPERATOR 1
using namespace std;
template <typename T>
ostream &operator<<(ostream &cout, vector<T> vec)
{
	cout << "{";
	for (int i = 0; i < (int)vec.size(); i++)
	{
		cout << vec[i];
		if (i != (int)vec.size() - 1)
			cout << ',';
	}
	cout << "}";
	return cout;
}

template <typename T1, typename T2>
ostream &operator<<(ostream &cout, pair<T1, T2> p)
{
	cout << "(" << p.first << ',' << p.second << ")";
	return cout;
}

template <typename T, typename T2>
ostream &operator<<(ostream &cout, set<T, T2> s)
{
	vector<T> t;
	for (auto x : s)
		t.push_back(x);
	cout << t;
	return cout;
}

template <typename T, typename T2>
ostream &operator<<(ostream &cout, multiset<T, T2> s)
{
	vector<T> t;
	for (auto x : s)
		t.push_back(x);
	cout << t;
	return cout;
}

template <typename T>
ostream &operator<<(ostream &cout, queue<T> q)
{
	vector<T> t;
	while (q.size())
	{
		t.push_back(q.front());
		q.pop();
	}
	cout << t;
	return cout;
}

template <typename T1, typename T2, typename T3>
ostream &operator<<(ostream &cout, map<T1, T2, T3> m)
{
	for (auto &x : m)
	{
		cout << "Key: " << x.first << ' ' << "Value: " << x.second << endl;
	}
	return cout;
}

template <typename T1, typename T2>
void operator+=(pair<T1, T2> &x,const pair<T1, T2> y)
{
	x.first += y.first;
	x.second += y.second;
}
template <typename T1,typename T2>
pair<T1,T2> operator + (const pair<T1,T2> &x,const pair<T1,T2> &y)
{
	return make_pair(x.first+y.first,x.second+y.second);
}

template <typename T1,typename T2>
pair<T1,T2> operator - (const pair<T1,T2> &x,const pair<T1,T2> &y)
{
	return mp(x.first-y.first,x.second-y.second);
}

template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> x)
{
	return make_pair(-x.first, -x.second);
}

template <typename T>
vector<vector<T>> operator~(vector<vector<T>> vec)
{
	vector<vector<T>> v;
	int n = vec.size(), m = vec[0].size();
	v.resize(m);
	for (int i = 0; i < m; i++)
	{
		v[i].resize(n);
	}
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			v[i][j] = vec[j][i];
		}
	}
	return v;
}
#endif
#include <sstream>
void print0x(int x)
{
	std::vector <int> vec;
	while(x)
	{
		vec.push_back(x&1);
		x>>=1;
	}
	std::reverse(vec.begin(),vec.end());
	for(int i=0;i<(int)vec.size();i++)
	{
		std::cout<<vec[i];
	}
	std::cout<<' ';
}

template <typename T>
void print0x(T x,int len)
{
	std::vector <int> vec;
	while(x)
	{
		vec.push_back(x&1);
		x>>=1;
	}
	reverse(vec.begin(),vec.end());
	for(int i=(int)vec.size();i<len;i++)
	{
		putchar('0');
	}
	for(size_t i=0;i<vec.size();i++)
	{
		std::cout<<vec[i];
	}
	std::cout<<' ';
}
vector<string> vec_splitter(string s) {
	s += ',';
	vector<string> res;
	while(!s.empty()) {
		res.push_back(s.substr(0, s.find(',')));
		s = s.substr(s.find(',') + 1);
	}
	return res;
}
void debug_out(
vector<string> __attribute__ ((unused)) args,
__attribute__ ((unused)) int idx, 
__attribute__ ((unused)) int LINE_NUM) { cerr << endl; } 
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
	if(idx > 0) cerr << ", "; else cerr << "Line(" << LINE_NUM << ") ";
	stringstream ss; ss << H;
	cerr << args[idx] << " = " << ss.str();
	debug_out(args, idx + 1, LINE_NUM, T...);
}
#define debug(...) debug_out(vec_splitter(#__VA_ARGS__), 0, __LINE__, __VA_ARGS__)
#endif
using namespace std;
int n,m;
char c[505][505];

void solve()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
	{
		scanf("%s",c[i]);
	}
	for(int i=0;i<n;i+=3)
	{
		for(int j=0;j<m;j++) c[i][j]='X';
	}
	for(int i=3;i<n;i+=3)
	{
		int flag=false;
		for(int z=i-2;z<i;z++)
		{
			for(int j=0;j<m;j++)
			{
				if(c[z][j]=='X' && !flag)
				{
					c[i-1+i-2-z][j]='X';
					flag=true;
				}
			}
		}
		if(!flag)
		{
			c[i-2][0]=c[i-1][0]='X';
		}
	}
	if(n%3==0)
	{
		for(int j=0;j<m;j++)
		{
			if(c[n-2][j]=='X' || c[n-1][j]=='X')
			{
				c[n-2][j]='X';
				c[n-1][j]='X';
			}
		}
	}
	for(int i=0;i<n;i++) printf("%s\n",c[i]);
}

int main()
{
//	// freopen("input.txt","r",stdin);
	int T;
	scanf("%d",&T);
	while(T--) solve();
	return 0;
}