“…In an identical way, we find all remaining cycles of the permutations π₯ and π¦ and given as: π₯ = (π 1 , π 126 )(π 2 , π 125 )(π 3 , π 124 )(π 4 , π 123 )(π 5 , π 122 )(π 6 , π 121 )(π 7 , π 120 )(π 8 , π 119 )(π 9 , π 118 ) (π , π 117 )(π 11 , π 116 )(π 12 , π 115 )(π 13 , π 114 )(π 14 , π 113 )(π 15 , π 112 )(π 16 , π 111 )(π 17 , π 110 )(π 18 , π 109 ) (π , π 108 )(π 20 , π 107 )(π 21 , π 106 )(π 22 , π 105 )(π 23 , π 104 )(π 24 , π 103 )(π 25 , π 102 )(π 26 , π 101 )(π 27 , π 100 ) (π 28 , π 99 )(π 29 , π 98 )(π 30 , π 97 )(π 31 , π 96 )(π 32 , π 95 )(π 33 , π 94 )( π 34 , π 93 )( π 35 π 92 )(π 36 , π 91 )(π 37 , π 90 ) (π 38 , π 89 )(π 39 , π 88 )(π 40 , π 87 )(π 41 , π 86 )(π 42 , π 85 )(π 43 , π 84 )(π 44 64 ). π¦ = (0 , β, 1)(π 1 , π 96 , π 30 ) (π 2 , π 65 , π 60 )(π 3 , π 120 , π 4 )(π 5 , π 45 , π 77 )(π 6 , π 113 , π 8 )(π 7 , π 124 , π 123 ) (π , π 83 , π 35 )(π 10 , π 90 , π 27 )(π 11 , π 93 , π <...…”