A. Hayato and Schooltime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputToday 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.InputThe 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.OutputFor 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.ExampleInput
631 1 141 1 2 231 2 351 4 5 1 242 6 2 455 6 3 2 1Output
YES
1 2 3
YES
3 4 1
NO
YES
1 3 4
NO
YES
1 3 5
NoteIn 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.
631 1 141 1 2 231 2 351 4 5 1 242 6 2 455 6 3 2 1
YES 1 2 3 YES 3 4 1 NO YES 1 3 4 NO YES 1 3 5 
['constructive algorithms', 'greedy', '*800']