CF 1780A: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 ai+aj+ak is odd.

The odd numbers are integers that are not divisible by 2: 1, 3, 5, and so on.

Input

The first line contains a single integer t (1≤t≤104) — the number of test cases.

For each test case, the first line contains one integer n (3≤n≤300) — the length of a

The second line contains n integers a1,a2,…,an (1≤ai≤105) — the array a

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105

Output

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≤i,j,k≤n) — the indices of the numbers. If there are several answers, output any.

Example

input

6

3

1 1 1

4

1 1 2 2

3

1 2 3

5

1 4 5 1 2

4

2 6 2 4

5

5 6 3 2 1

output

YES

1 2 3

YES

3 4 1

NO

YES

1 3 4

NO

YES

1 3 5

Note

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.

找到3个索引值使得这个数组的和是奇数，如果不是输出NO，如果是输出yes跟对应的索引值

如果是奇数，要么三个都是奇数，要么一个奇数，2个偶数；

依次判断即可：

下面是代码：