“…, f (m − 1)). We denote by ord the Newton polygon of G 0,1 ⊕ G 1,0 which has slopes 0 and 1 with multiplicity 1 and by ss the Newton polygon of G 1,1 which has slope 1/2 with multiplicity 2. m = 3 p 1 mod 3 2 mod 3 prime orbits split (1, 2) a signature (1, 1, 1 1,1,0,0,0,0) ord 3 ord 2 ⊕ ss ord ⊕ ss 2 ss 3 (1,2,5) (1,1,0,0,1,0,0) ord 3 ss 3 ord 3 ss 3 (1,3,4) (1,0,1,0,0,0,0) ord 2 ord 2 ss 2 ss 2 m = 9 p 1 mod 9 2, 5 mod 9 4, 7 mod 9 8 mod 9 (1,2,4,8,7,5) (1, 4, 7), (2,8,5) (1, 8), (2, 7) prime orbits split (3,6) (3), (6) (4, 5), (3, 6) a signature (1,1,7) (1,1,1,1,0,0,0,0) ord 4 ss 4 (1/3, 2/3) ⊕ ord ss 4 (1,2,6) (1,1,0,0,1,0,0,0) ord 3 ss 3 (1/3, 2/3) ss 3 (1,3,5) (1,1,0,1,0,0,0,0) ord 3 ss 3 (1/3, 2/3) ss 3 m = 10 p 1 mod 10 3, 7 mod 10 9 mod 10 (1, 3,9,7) (1, 9), (2, 8) prime orbits split (2,6,8,4), (5) (3, 7), (4, 6), (5) a signature (1,1,8) (1,1,1,1,0,0,0,0,0) ord 4 ss 4 ss 4 (1,2,7) (1,1,1,0,0,1,0,0,0) ord 4 ss 4 ss 4 (1,4,5) (1,0,1,0,0,0,0,0,0) ord 2 ss 2 ss 2 m = 11 p 1 mod 11 2, 6, 7, 8 mod 11 3, 4, 5, 9 mod 11 10 mod 11 (1,3,4,5,9) (1, 10), (2,9) (3,9) (1, 11), (4, 8) (1, 5), (2, 10), (3) (2), (4), (5,11) (3,9), (2, 10) prime orbits split (4, 8), (6), (7, 11), (9) (6), (8), (10) (5, 7), (6) a signature (1,1,10) (1,1,1,1,1,0,0,0,0,0,0) ord 5 ord 3 ⊕ ss 2 ord 2 ⊕ ss 3 ss 5 (1,2,9) (1,1,1,0,0,0,1,0,0,0,0) ord 4 ord ⊕ ss 3 ord 3 ⊕ ss ss 4 (1,3,8) (1,1,0,0,1,0,0,0,0,0,0) ord 3 ord 2 ⊕ ss ord ⊕ ss 2 ss 3 (1,4,7) (1,1,0,1,0,0,1,0,0,0,0) ord 4 ss 4 ord 4 ss 4 (1...…”