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
                        chars = [s[i], s[j], s[k], s[l], s[m]]
                        
                        # CPCTF的かどうかを判定
                        # 1. 1文字目と3文字目が等しい
                        if chars[0] == chars[2]:
                            # 2. それ以外の文字の組み合わせはすべて異なる
                            if chars[0] != chars[1] and chars[0] != chars[3] and chars[0] != chars[4] and \
                               chars[1] != chars[2] and chars[1] != chars[3] and chars[1] != chars[4] and \
                               chars[2] != chars[3] and chars[2] != chars[4] and \
                               chars[3] != chars[4]:
                                count += 1
    
    return count