Displaying the button at the button after the two tables

[Persi]

I have two tables in one form and a button in another form. So the two tables must be displayed first and then the button. But for me the button is displayed between the two tables. Can anyone help me??

This is my code

<form action="choirEvedit.php5" method="post">
<?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("centreformusic", $con);
   $res=mysql_query("select * from choirevent where ProgId='$_GET[pid]' and event_date='$_GET[dt]'") or die(mysql_error());
        echo "Events<br><br>";
        echo "<input type='hidden' name='pid' value='$_GET[pid]'>";
        echo "<table border='0' cellpadding='3' cellspacing='3' class='name'>";
        echo "<tr><th>Event Date:</th><th>Time:</th><th>Title:</th></tr>";
        while($row=mysql_fetch_array($res))
        {   
          
           echo "<tr><td><a href='choirEved.php5?pid=".$row[ProgId]."&dt=".$row[event_date]."'>".$row[event_date]."</a></td>";
           echo "<td>".$row[Time]."</td>";
           echo "<td>".$row[title]."</td></tr>"; 
        }
        echo "</table>";
  $sql=mysql_query("select * from song where ProgId='$_GET[pid]'") or die(mysql_error());
   echo "<table border='0' cellpadding='3' cellspacing='3' class='name'>";   
   $count=0;
   echo "<tr><th>S.No.</th><th>Song</th><th>Author</th>";
   while($row=mysql_fetch_array($sql))
   {
       $count=$count+1;
       echo "<tr><td><a href='choirEvSong.php5?pid=".$row[ProgId]."&sid=".$row[SongID]."'>$count</a></td>";
       echo "<td>".$row[Song]."</td>";
       echo "<td>".$row[Author]."</td></tr>";
   }
  
mysql_close($con);
?>
</form>

<form action="choirEvent1.php5" method="post">
<input type="image" src="images/back.png">
</form>