Session variable and Dynamic Pagination in php - php

Question

<form action='movies.php' method='POST'>
Language: <select name="language">
<option selected>hindi</option>
<?php
require("config.php");
$result="SELECT language FROM movies";
$q = mysqli_query($conn,$result) or die(mysql_error());
while ($row=mysqli_fetch_array($q)) {
$s1=$row["language"];
echo "<option>
$s1
</option>";
}
echo "<br>"
?>
</select>
<br /> <br />
<input type='submit' value='Submit' />
</form>
<?php
$lang=#$_POST['language'];
$_SESSION["lang1"]=$lang;
/*
if($lang){
$sql = "SELECT name,language FROM movies WHERE language='$lang'";
$result = mysqli_query($conn,$sql);
if (mysqli_num_rows($result)>0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<div class="query"> <img src="images\tiles\bb.jpg"> ';
echo "<h3> " . $row["name"]."</h3>". "<br>";
echo "</div>";
}
} else {
echo "0 results";
}
}
*/
$page=#$_GET['page'];
if($page==""|| $page=="1")
{
$page1=0;
}
else{
$page1=($page*3)-3;
}
$sql = "SELECT name,language FROM movies WHERE language='". $_SESSION['lang1']."' limit $page1,3 ";
$result = mysqli_query($conn,$sql);
$a= mysqli_num_rows($result);
while ($list=mysqli_fetch_array($result))
{
echo $list['name'] . " : " . $list['language'] . "<br />";
}
$sql = "SELECT name,language FROM movies WHERE language='$lang'";
$result = mysqli_query($conn,$sql);
$a= mysqli_num_rows($result);
$numrows=$a;
$rowsperpage=3;
$totalpages= ceil($numrows/$rowsperpage);
echo "</br>";
for($b=1;$b<=$totalpages;$b++)
{
?><?php echo $b." ";?> <?php
}
?>
I get proper output on movies.php which is the first 3 rows from database,but when i click on the dynamically created pagination link like movies.php?page=2 then there is no output on this page.This code works fine if i manually set the
$_SESSION["lang1"]="English"; then i get proper output but when i take input from the form it doesnt work.