If \( \gcd(m,n)=1 \) and \( 1^2-2^2+3^2-4^2+\ldots+(2021)^2-(2022)^2+(2023)^2=1012\, m^2 n \) then \( m^2-n^2 \) is equal to:
\( 240 \)
\( 200 \)
\( 220 \)
\( 180 \)