/*

     题意：问最少替换'*'为'.'，使得'.'连通的都是矩形

     BFS：搜索想法很奇妙，先把'.'的入队，然后对于每个'.'八个方向寻找

         在2*2的方格里，若只有一个是'*'，那么它一定要被替换掉

 */

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <queue>

 using namespace std;

 const int MAXN = 2e3 + ;

 const int INF = 0x3f3f3f3f;

 int n, m;

 int dx[][] = {{,,},{,-,-},{-,-,},{,,}};

 int dy[][] = {{,,},{,,},{,-,-},{,-,-}};

 char s[MAXN][MAXN];

 bool ok(int x, int y)

 {

     if (x <  || x >= n)    return false;

     if (y <  || y >= m)    return false;

     return true;

 }

 void BFS(void)

 {

     queue<pair<int, int> > Q;

     for (int i=; i<n; ++i)

     {

         for (int j=; j<m; ++j)

         {

             if (s[i][j] == '.')

             {

                 Q.push (make_pair (i, j));

             }

         }

     }

     while (!Q.empty ())

     {

         int x = Q.front ().first;    int y = Q.front ().second;

         Q.pop ();

         for (int i=; i<; ++i)

         {

             int cnt = ;    int px, py;    bool flag = true;

             for (int j=; j<; ++j)

             {

                 int tx = x + dx[i][j];    int ty = y + dy[i][j];

                 if (ok (tx, ty))

                 {

                     if (s[tx][ty] == '*')

                     {

                         cnt++;    px = tx;    py = ty;

                     }

                 }

                 else    flag = false;

             }

             if (flag && cnt == )

             {

                 s[px][py] = '.';    Q.push (make_pair (px, py));

             }

         }

     }

 }

 int main(void)        //Codeforces Round #297 (Div. 2) D. Arthur and Walls

 {

     while (scanf ("%d%d", &n, &m) == )

     {

         for (int i=; i<n; ++i)    scanf ("%s", s[i]);

         BFS ();

         for (int i=; i<n; ++i)    printf ("%s\n", s[i]);

     }

     return ;

 }

 /*

 5 5

 .*.*.

 *****

 .*.*.

 *****

 .*.*.

 6 7

 ***.*.*

 ..*.*.*

 *.*.*.*

 *.*.*.*

 ..*...*

 *******

 */