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A15764. Hayato and School

编程题 普及/提高-

题目描述

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.

输入格式

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 $2\cdot10^5$ .

输出格式

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.

输入输出样例

输入 #1
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
输出 #1
YES
1 2 3
YES
3 4 1
NO
YES
1 3 4
NO
YES
1 3 5

说明/提示

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.
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