Heptomino Rectification Puzzle

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Puzzle description: The problem of polyomino rectification is to fill a rectangle with copies of a single polyomino. For some of the polyominoes the solution with the mininmal number of pieces needs a quite large rectangle and is surprisingly difficult. For the P-heptomino considered here the minimal rectangle has size 14x14 and consists of 28 identical P-heptominoes. Each heptomino consists of seven solid cubies. So you also can try to built cuboids. Try to use some of the P-heptominoes to fill the following boxes: 3x3x7, 3x4x7, 3x5x7, ... 4x4x7, 4x5x7, 4x6x7, ... 5x5x7, 4x6x7, 4x7x7, ... With more pieces you can fill any box having one side divisible by 7. Source: https://www.math.uni-bielefeld.de/~sillke/PENTA/qu7-p3 If you can read German: Mehr Informationen gibt es in https://welt-der-geduldspiele.blogspot.com/search?q=3D-Druck Puzzle difficulty: The 14x14 square is of advanced difficulty. If you do not succeed immediately, the tray contains a spare place for one heptomino. Printing instructions: Print the tray and 18 pieces of the G-hexomino. Prints are fine with no supports. I printed the puzzle in PLA at 0.1 mm layer height.