#P1780A. Hayato and School
Hayato and School
Description
Today Hayato came home from school with homework.
In the assignment, Hayato was given an array of length . The task was to find numbers in this array whose sum is odd. At school, he claimed that there are such numbers, but Hayato was not sure, so he asked you for help.
Answer if there are such three numbers, and if so, output indices , , and such that is odd.
The odd numbers are integers that are not divisible by : , , , and so on.
The first line contains a single integer () — the number of test cases.
For each test case, the first line contains one integer () — the length of .
The second line contains integers () — the array .
It is guaranteed that the sum of over all test cases does not exceed .
For each test case, in the first line print one word "YES" (without quotes) if there are numbers with an odd sum or "NO" (without quotes) if there are no such numbers.
If the answer exists, then on the second line print distinct integers () — the indices of the numbers. If there are several answers, output any.
Input
The first line contains a single integer () — the number of test cases.
For each test case, the first line contains one integer () — the length of .
The second line contains integers () — the array .
It is guaranteed that the sum of over all test cases does not exceed .
Output
For each test case, in the first line print one word "YES" (without quotes) if there are numbers with an odd sum or "NO" (without quotes) if there are no such numbers.
If the answer exists, then on the second line print distinct integers () — the indices of the numbers. If there are several answers, output any.
Note
In the first test case, there is one way to choose numbers, and since , this triple is fine for us.
In the second test case, you need to choose the numbers , since .
In the third test case, there is one way to choose three numbers, but 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.