“…can be derived as the following cases: Case 1: if N 1 is even. D WNA + can be expressed asD WNA + = {αn 11 d + βn 2 d, n 2 ∈ [0, N 2 − 1], n 11 = 0, 2, × …, N 1 } ∪ {αn 12 d + βn 2 d, n 2 ∈ [0, N 2 − 1], n 12 = 1, 3, …, N 1 − 1} ∪ {αN 1 d + β(N 2 − 1)d + αn 3 d, n 3 = 1, 3, …, N 1 − 1} (24) Assume that D E + = {αN 1 d + β(N 2 − 1)d + αn 3 d, n 3 = 1, 3, ⋯, N 1 − 1}, by combining(23) and(24), we can conclude that D E + contains N 1 /2 unique lags and satisfies the condition ofD d + βn 2 d, n 2 ∈ [0, N 2 − 1], n 11 = 0, 2, × …, N 1 − 1} ∪ {αn 12 d + βn 2 d, n 2 ∈ [0, N 2 − 1], × n 12 = 1, 3, …, N 1 } ∪ {αN 1 d + β(N 2 − 1)d + αn 3 d, × n 3 = 2, 4, …, N 1 − 1} 1 d + β(N 2 − 1)d + αn 3 d, n 3 = 2, 4, …, N 1 − 1}.It is obvious that D E + has N 1 /2 unique lags and D (according to(23) and(25)). …”