To evaluate the performance of our first year CS majored students,

为了计算第一年计算机科学学生的表现

we consider their grades of three courses only:

我们从三个方面考虑我们的成绩

C - C Programming Language, M - Mathematics (Calculus or Linear Algebra), and E - English.

C语言,数学,和英语

At the mean time, we encourage students by emphasizing on their best ranks --

同时我们鼓励我们的学生强调自己的排名

that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

就是,通过四个排名包含三门课和平均数,我们打印出每个学生最好的排名

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

例如,四个成绩CMEA。

StudentID  C  M  E  A
310101 98 85 88 90
310102 70 95 88 84
310103 82 87 94 88
310104 91 91 91 91

Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

第一名是C 语言成绩第一的,第二名是数学第一的,第三名是英语第一的,最后一名是平均分第一的

Input

Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.

Output

For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.

The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.

If a student is not on the grading list, simply output "N/A".

Sample Input

5 6
310101 98 85 88 90
310102 70 95 88 84
310103 82 87 94 87
310104 91 91 91 91
310105 85 90 90 88
310101
310102
310103
310104
310105
999999

Sample Output

1 C
1 M
1 E
1 A
3 A
N/A

简单题,下面不想翻译了,思路为四次排序,然后保存四次的排名,然后通过学号找到这四次的排名,按照A > C > M > E,输出好的排名就行了,没有学号输出N/A

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