def count_cpctf_substrings(s):
    n = len(s)
    count = 0
    
    # 5つの位置を選ぶすべての組み合わせを考える
    for i in range(n):
        for j in range(i+1, n):
            for k in range(j+1, n):
                for l in range(k+1, n):
                    for m in range(l+1, n):
                        # 選んだ5文字
                        sub = [s[i], s[j], s[k], s[l], s[m]]
                        
                        # CPCTF的かどうかを判定
                        # 1文字目と3文字目が等しい
                        if sub[0] == sub[2]:
                            # それ以外の文字はすべて異なる
                            if (sub[0] != sub[1] and sub[0] != sub[3] and sub[0] != sub[4] and
                                sub[1] != sub[2] and sub[1] != sub[3] and sub[1] != sub[4] and
                                sub[2] != sub[3] and sub[2] != sub[4] and
                                sub[3] != sub[4]):
                                count += 1
    
    return count