Quack the Angry Duck No one is quite certain what the emoticon &gt;:U is intended to represent, but many scholars believe it looks like an angry duck. Let's assume that's the case.

Task

Given an integer n between 0 and 3 inclusive, print or return

quack


if n = 0,

&gt;:U


if n = 1,

     U   U
&gt;  : U   U
 &gt;   U   U
&gt;  : U   U
      UUU


if n = 2, or

                  &gt;:U         &gt;:U
&gt;:U               &gt;:U         &gt;:U
   &gt;:U       &gt;:U  &gt;:U         &gt;:U
      &gt;:U         &gt;:U         &gt;:U
         &gt;:U      &gt;:U         &gt;:U
      &gt;:U         &gt;:U         &gt;:U
   &gt;:U       &gt;:U  &gt;:U         &gt;:U
&gt;:U               &gt;:U         &gt;:U
                     &gt;:U&gt;:U&gt;:U


if n = 3.

You can assume the input will always be valid. There should be no leading space in the output, but any amount of trailing space is fine. Ducks (with the possible exception of @cobaltduck) have no tolerance for loopholes. Shortest code in bytes wins.
Java, 303 286 bytes

Saved 17 bytes thanks to @VoteToClose!

Not actually meant to be the shortest, I just thought it would be fun to try in Java.

Creates a list of string variables representing common strings, then creates an array of all the outputs, then prints the right one.

String a(int y){String n="\n",d="&gt;:U",A=" ",B=A+A,C=B+B,D=C+C,a="U"+B+" U"+n,G=D+A,H=C+B,c=d+G+d+n,E=B+A,F=C+E;String[]z={"quack",d,C+A+a+"&gt;"+B+": "+a+" &gt;"+E+a+"&gt;"+B+": "+a+C+B+"UUU",D+D+B+c+d+D+F+c+B+A+d+F+d+B+c+H+d+G+c+G+d+H+c+H+d+G+c+E+d+F+d+B+c+d+D+F+c+D+D+C+A+d+d+d};return z[y];}


Ungolfed:

String a(int y) {
    String n = "\n", d = "&gt;:U", A = " ", B = A + A, C = B + B, D = C + C,
            a = "U" + B + " U" + n, G = D + A, H = C + B, c = d + G + d + n,
            E = B + A, F = C + E;
    String[] z = { "quack", d, C + A + a + "&gt;" + B + ": " + a + " &gt;" + E + a + "&gt;" + B + ": " + a + C + B + "UUU", D + D + B + c + d + D + F + c + B + A + d + F + d + B + c + H + d + G + c + G + d + H + c + H + d + G + c + E + d + F + d + B + c + d + D + F + c + D + D + C + A + d + d + d };
    return z[y];
}


This is my first answer on this site, so please tell me if I have done something wrong.