B. All Distinct

(大家好，拖更了这么久，本博主又回来了)。

B. All Distinct

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Sho has an array aa consisting of nn integers. An operation consists of choosing two distinct indices ii and jj and removing aiai and ajaj from the array.

For example, for the array [2,3,4,2,5] Sho can choose to remove indices 1 and 3. After this operation, the array becomes [3,2,5]. Note that after any operation, the length of the array is reduced by two.

After he made some operations, Sho has an array that has only distinct elements. In addition, he made operations such that the resulting array is the longest possible.

More formally, the array after Sho has made his operations respects these criteria:

• No pairs such that (i<j) and ai=aj exist.
• The length of aa is maximized.

Output the length of the final array.

Input

The first line contains a single integer tt (1≤t≤103) — the number of test cases.

The first line of each test case contains a single integer nn (1≤n≤50) — the length of the array.

The second line of each test case contains nn integers aiai (1≤ai≤104) — the elements of the array.

Output

For each test case, output a single integer — the length of the final array. Remember that in the final array, all elements are different, and its length is maximum.

Example

input

Copy

4
6
2 2 2 3 3 3
5
9 1 9 9 1
4
15 16 16 15
4
10 100 1000 10000

output

Copy

2
1
2
4

Note

For the first test case Sho can perform operations as follows:

1. Choose indices 1 and 5 to remove. The array becomes [2,2,2,3,3,3]→[2,2,3,3].
2. Choose indices 1 and 4 to remove. The array becomes [2,2,3,3]→[2,3].

The final array has a length of 22, so the answer is 22. It can be proven that Sho cannot obtain an array with a longer length.

For the second test case Sho can perform operations as follows:

1. Choose indices 3 and 4 to remove. The array becomes [9,1,9,9,1]→[9,1,1].
2. Choose indices 1 and 3 to remove. The array becomes [9,1,1]→[1].

The final array has a length of 1, so the answer is 1. It can be proven that Sho cannot obtain an array with a longer length.

首先我们要知道，我们需要找到一个最大长度且数组元素不重复的数组。

接着我们把数组分区，将每个数组元素的个数重建一个数组。

如果该数组元素的数量是奇数个，我们就置位1

如果该数组元素的数量是偶数个，我们就置位2

那么我就开始讨论数组中元素的等可能的情况。

（注讨论的情况，重复是指该数组值出现等于两次，数组数量为1的情况，我们直接加就完事了）

长度默认为0

1.数组中元素全不重复

2数组中元素有偶数个重复

例如 1 1 2 2

等于 1 2

长度加有多少个重复

3.数组中元素有奇数个重复

例如 1 1 2 2 3 3

等于 1 2

长度加有多少个重复-1

思路已经分析完了，接下来我直接贴一下代码。

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t > 0)
    {
        int n, index1 = 0, sum = 0;
        cin >> n;
        int data1[10000];
        int qu[20000];
        for (int i = 0; i < n; i++)
        {
            cin >> data1[i];
        }
        sort(data1, data1 + n);
        for (int i = 0; i <= data1[n - 1]; i++)
        {
            qu[i] = 0;
        }
        for (int i = 0; i < n; i++)
        {
            qu[data1[i]]++;
        }
        for (int i = 0; i <= data1[n - 1]; i++)
        {
            if (qu[i] != 0)
            {
                if (qu[i] % 2 == 0)
                {
                    qu[i] = 2;
                }
                else
                {
                    qu[i] = 1;
                }
            }
        }
        for (int i = 1; i <= data1[n - 1]; i++)
        {
            if (qu[i] == 2)
            {
                index1++;
            }
        }
        if (index1 != 0)
        {
            if (index1 % 2 == 0)
            {
                sum = index1;
            }
            else
            {
                index1 = index1 - 1;
                sum = index1;
            }
            for (int i = 1; i <= data1[n - 1]; i++)
            {
                if (qu[i] == 1)
                {
                    sum++;
                }
            }
            cout << sum << endl;
        }
        else if (index1 == 0)
        {
            for (int i = 1; i <= data1[n - 1]; i++)
            {
                if (qu[i] == 1)
                {
                    sum++;
                }
            }
            cout << sum << endl;
        }
        t--;
    }
    return 0;
}