Starting with a standard double-six set of dominos, arrange the 28 dominos into the perimeter of a square, such that the pip values on each of the four sides of the square sum to the same value. At each corner of the square, there is a 90^o turn, so that each side is 7½ dominos long.

The usual rules apply; that is, where two dominos meet, the ends must match (have the same number of pips).

Place the double value dominos lengthwise, not crosswise. Present your answer as four strings of 7½ dominos each, starting at the upper left corner and proceeding clockwise around the square. For example, the top edge might be: 0-3|3-3|3-5|5-5|5-0|0-2|2-4|4 (sum = 44), and the right edge might be: 4-6|6-2|2-2|2-1|1-5|5-4|4-0|0 (sum = 44). There are multiple answers, and any valid solution will be accepted.