from itertools import combinations

def count_cpctf_like_subsequences(S):
    N = len(S)
    count = 0
    for i, j, k, l, m in combinations(range(N), 5):
        s = [S[i], S[j], S[k], S[l], S[m]]  # s1, s2, s3, s4, s5
        if s[0] == s[2]:  # s1 == s3
            # i = j or {i,j} = {1,3} でのみ等しい ⇒ 他の組み合わせはすべて異なる
            if len(s) == len(set(s)) + 1:  # 同じ文字が1組だけ（s1==s3）のとき、重複は1つだけ
                if s[1] != s[3] and s[1] != s[4] and s[3] != s[4]:  # s2, s4, s5 も違うことを明示
                    count += 1
    return count