linux - Regarding the mixed use of while read and sed, sed eats the first line of standard input

There is a crashcourse_test.txt file with the following content:

# cat crashcourse_test.txt 
1001|Anvils R Us|123 Main Street|Southfield|MI|48075|USA
1002|LT Supplies|500 Park Street|Anytown|OH|44333|USA
1003|ACME|555 High Street|Los Angeles|CA|90046|USA
1004|Furball Inc.|1000 5th Avenue|New York|NY|11111|USA
1005|Jet Set|42 Galaxy Road|London|NULL|N16 6PS|England
1006|Jouets Et Ours|1 Rue Amusement|Paris|NULL|45678|France
# 

The content of the while_read.sh script is as follows. The purpose is to use the above file content as standard input, pass it into the while read loop, and replace the "USA" string in each line with "XXXXXXXXX" through the sed command:

#!/bin/bash

while read line
do
        sed 's/USA/XXXXXXXXX/'
done < crashcourse_test.txt

The output result is:

# ./while_read.sh 
1002|LT Supplies|500 Park Street|Anytown|OH|44333|XXXXXXXXX
1003|ACME|555 High Street|Los Angeles|CA|90046|XXXXXXXXX
1004|Furball Inc.|1000 5th Avenue|New York|NY|11111|XXXXXXXXX
1005|Jet Set|42 Galaxy Road|London|NULL|N16 6PS|England
1006|Jouets Et Ours|1 Rue Amusement|Paris|NULL|45678|France
# 

You can see that the line 1001 contains the string "USA" ( happens to be the first line in the file ), which meets the replacement requirements, but the result is that this line is not output.
I know the way to output it correctly, which can be achieved by using echo $line | sed 's/USA/XXXXXXXXX/'.
But I want to know the reason why only the first line of output is missed. Where does the first line go after entering the while read as standard input?