
148  The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple  B
It's time to prepare the problems for the 14th Zhejiang Provincial Collegiate Programming Contest! Almost all members of Zhejiang University programming contest problem setter team brainstorm and code day and night to catch the deadline, and empty bottles of Marjar Cola litter the floor almost everywhere! To make matters worse, one of the team member fell ill just before the deadline. So you, a brilliant student, are found by the team leader Dai to help the team check the problems' arrangement. Now you are given the difficulty score of all problems. Dai introduces you the rules of the arrangement:
The team members have given you lots of possible arrangements. Please check whether these arrangements obey the rules or not. InputThere are multiple test cases. The first line of the input is an integer T (1 ≤ T ≤ 10^{4}), indicating the number of test cases. Then T test cases follow. The first line of each test case contains one integer n (1 ≤ n ≤ 100), indicating the number of problems. The next line contains n integers s_{1}, s_{2}, ... , s_{n} (1000 ≤ s_{i} ≤ 1000), indicating the difficulty score of each problem. We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++. OutputFor each test case, output "Yes" (without the quotes) if the arrangement follows the rules, otherwise output "No" (without the quotes). Sample Input8 9 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 11 999 1 1 2 3 4 5 6 7 8 9 11 999 1 3 5 7 9 11 13 17 19 21 10 15 1 13 17 1 7 9 5 3 11 13 1 1 1 1 1 1 1 1 1 1 1 1 2 10 2 3 4 5 6 7 8 9 10 11 10 15 1 13 3 6 5 4 7 1 14 Sample OutputNo No Yes No Yes Yes No No HintThe first arrangement has 9 problems only, which violates the first rule. Only one problem in the second and the fourth arrangement has a difficulty score of 1, which violates the third rule. The easiest problem in the seventh arrangement is a problem with a difficulty score of 2, which violates the second rule. After sorting the problems of the eighth arrangement by their difficulty scores in ascending order, we can get the sequence 1, 1, 3, 4, 5, 6, 7, 13, 14, 15. We can easily discover that 13  7 = 6 > 2. As the problem with a difficulty score of 13 is not the hardest problem (the hardest problem in this arrangement is the problem with a difficulty score of 15), it violates the fourth rule. 