Printing first four lines per key

Question

Dear all

I would like to print the first 4 lines by field 1 using awk

Input

111 1032192
111 2323476
111 1698881
111 2451712
111 2013780
112 2331004
112 1886376
112 1189765
112 1877267

Output

111 1032192
111 2323476
111 1698881
111 2451712
112 2331004
112 1886376
112 1189765
112 1877267

Thanks

Tony

Answer 1

($1 != curkey) { curkey=$1; i=0; }
(i < 4)        { i++; print; }

Answer 2

awk '++count[$1] <= 4' input_file

Answer 3

Dear Glen and Larsmans,

Thank you so much for your prompt response. I have not tried the former suggestion, but the latter (with awk) works like a charm.

Cheers,

Tony