What percentage of users were using other browsers according to the source that reported 21.22% were using Firefox?
CREATE TABLE table_65881 (
    "Source" text,
    "Chrome" text,
    "Internet Explorer" text,
    "Firefox" text,
    "Safari" text,
    "Opera" text,
    "Other" text
)
SELECT "Other" FROM table_65881 WHERE "Firefox" = '21.22%'
Below are sql tables schemas paired with instruction that describes a task. Using valid SQLite, write a response that appropriately completes the request for the provided tables. ### Instruction: What percentage of users were using other browsers according to the source that reported 21.22% were using Firefox? ### Input: CREATE TABLE table_65881 (
    "Source" text,
    "Chrome" text,
    "Internet Explorer" text,
    "Firefox" text,
    "Safari" text,
    "Opera" text,
    "Other" text
) ### Response: SELECT "Other" FROM table_65881 WHERE "Firefox" = '21.22%'