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
                        substring = [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
                        # 2. それ以外の文字はすべて異なる
                        if substring[0] == substring[2] and \
                           substring[0] != substring[1] and \
                           substring[0] != substring[3] and \
                           substring[0] != substring[4] and \
                           substring[1] != substring[2] and \
                           substring[1] != substring[3] and \
                           substring[1] != substring[4] and \
                           substring[2] != substring[3] and \
                           substring[2] != substring[4] and \
                           substring[3] != substring[4]:
                            count += 1
    
    return count