CF1780A Hayato and School

Today Hayato came home from school with homework. In the assignment, Hayato was given an array $ a $ of length $ n $ . The task was to find $ 3 $ numbers in this array whose sum is odd. At school, he claimed that there are such $ 3 $ numbers, but Hayato was not sure, so he asked you for help. Answer if there are such three numbers, and if so, output indices $ i $ , $ j $ , and $ k $ such that $ a_i + a_j + a_k $ is odd. The odd numbers are integers that are not divisible by $ 2 $ : $ 1 $ , $ 3 $ , $ 5 $ , and so on.

The first line contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. For each test case, the first line contains one integer $ n $ ( $ 3 \le n \le 300 $ ) — the length of $ a $ . The second line contains $ n $ integers $ a_1, a_2, \ldots, a_n $ ( $ 1 \le a_i \le 10^5 $ ) — the array $ a $ . It is guaranteed that the sum of $ n $ over all test cases does not exceed $ 2\cdot10^5 $ .

For each test case, in the first line print one word "YES" (without quotes) if there are $ 3 $ numbers with an odd sum or "NO" (without quotes) if there are no such $ 3 $ numbers. If the answer exists, then on the second line print $ 3 $ distinct integers $ i, j, k $ ( $ 1 \le i, j, k \le n $ ) — the indices of the numbers. If there are several answers, output any.

In the first test case, there is one way to choose $ 3 $ numbers, and since $ 1 + 1 + 1 = 3 $ , this triple is fine for us. In the second test case, you need to choose the numbers $ 1, 2, 2 $ , since $ 1 + 2 + 2 = 5 $ . In the third test case, there is one way to choose three numbers, but $ 1 + 2 + 3 = 6 $ is an even number, so the required triple does not exist. In the fifth test case, no matter what three numbers we choose, their sum is even.