Rahul is a serial killer. Rahul has been betrayed by his lover in the past and now he want to eliminate entire Universe.He has already Eliminated majority of the population and now only a handful number of people are left. Like other Serial killers, he has an interesting pattern of killing people. 
He either kill one individual at a time or if he find two individuals of different heights,he eliminates both of them simultaneously. Now Rahul wants to eliminate them as quickly as he can.
So given $N$ as the number of people left and an array containing height of those $N$ people,tell the minimum number of kills Rahul require to eliminate the entire universe.

-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow. 
- Each test case constitutes of Two lines. 
- First line contains $N$, representing the number of people left in the universe
- The second line contains an array $a[i]$ of size $N$ containing heights of those $N$ people.

-----Output:-----
For each testcase, you have to output a Single line Containing the minimum number of kills required by Rahul to eliminate the Universe.

-----Constraints-----
- $1 \leq T \leq 50000$
- $1 \leq N \leq 50000$
- $100 \leq a[i] \leq 10^5$

-----Sample Input:-----
1
10
178 184 178 177 171 173 171 183 171 175

-----Sample Output:-----
5

Difficulty: interview

Starter code:

Solutions:
from math import ceil
t=int(input())
for i in range(t):
       p=int(input())
       l=list(map(int,input().split()))
       maxx=1
       for i in range(len(l)):
           maxx=max(maxx,l.count(l[i]))
       if(maxx*2>p):
           print(maxx)
       else:
           q=p-maxx*2
           maxx+=ceil(q/2)
           print(maxx)
           
       
              



#dt = {} for i in x: dt[i] = dt.get(i,0)+1
import sys;input = sys.stdin.readline
inp,ip = lambda :int(input()),lambda :[int(w) for w in input().split()]

for _ in range(inp()):
    n = inp()
    x = ip()
    dt = {} 
    for i in x: dt[i] = dt.get(i,0)+1
    ans = max(max(dt.values()),(n-1)//2+1)
    print(ans)