1265 card sum

Description

After watching the performances, I quickly understood the secrets behind the magic. Then I came up with a similar magic. He asked his father to extract n cards from the 10000 cards, and then he would say the number of M cards at random, you can quickly determine whether the number is the sum of one or two of the N cards.
Of course, as the saying goes, if you have a father, you must have his son. Quick Bi is also an outstanding software engineer.

Input

An integer in the first row indicates the number of groups (multiple groups of data). For each group of data, the first row has two integers n (1 ≤ n ≤ 10000 ), M (1 ≤ m ≤ 100 ). The second row is the set of n different integers. The third row has m Integers to be verified.

Output

Output a line for each integer to be verified. If the set S contains two integers and the sum of the Integers to be verified, output "yes"; otherwise, output "no ".

Sample Input

110 53 1 8 4 7 9 5 10 2 627 7 26 10 11

Sample output

Noyesnoyesyes

Create an index with an array, and the number is 1 in the array. In this way, the subscript is used to determine whether the index is composed of two numbers, determine whether the array content is 1

 # include 
   
     main () {int number, Te; int I, J, K, M; int number1, number2; int A [10000]; int result [100000]; int B; int flag; scanf ("% d", & number); For (TE = 1; te <= number; Te ++) {for (k = 0; k <100000; k ++) result [k] = 0; scanf ("% d", & number1, & number2); for (I = 1; I <= number1; I ++) {scanf ("% d", & A [I]); Result [A [I] = 1 ;}for (I = 1; I <= number2; I ++) {flag = 0; scanf ("% d", & B); For (j = 1; j <= number1/2; j ++) {If (result [B-A [J] = 1) Flag = 1; elsecontinue;} If (flag = 1) printf ("Yes \ n"); else printf ("NO \ n") ;}}