You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times: 

  * you select two different numbers in the array a_i and a_j; 
  * you remove i-th and j-th elements from the array. 



For example, if n=6 and a=[1, 6, 1, 1, 4, 4], then you can perform the following sequence of operations: 

  * select i=1, j=5. The array a becomes equal to [6, 1, 1, 4]; 
  * select i=1, j=2. The array a becomes equal to [1, 4]. 



What can be the minimum size of the array after applying some sequence of operations to it?

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) is length of the array a.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output the minimum possible size of the array after applying some sequence of operations to it.

Example

Input


5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1


Output


0
0
2
1
0
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#include "debug.h"
#define deb(x...) cerr << "[" << #x << "] = ["; _print(x);
#else
#define deb(x...)
#endif
#define endl "\n"
#define pb push_back
#define int long long int
#define ll long long
#define pii pair<int,int>
#define F first
#define S second
#define mod 1000000007
#define MOD 998244353
#define inf (int)(1e18)
#define all(c) c.begin(),c.end()
#define read(v) for(auto &it:v) cin>>it;
#define print(v) for(auto &it:v) cout<<it<<" ";
#define getunique(a) a.resize(unique(a.begin(),a.end())-a.begin());
const int  N = 2e5 + 5;
void solve() {

	int n;
	cin >> n;

	vector <int> v(n);
	read(v);

	map <int, int> mpp;

	for (auto it : v) {
		mpp[it]++;
	}

	int mx = -1, element = -1;

	for (auto it : v) {
		if (mpp[it] > mx) {
			mx = mpp[it];
			element = it;
		}
	}

	mpp[element] = 0;

	int cur = mx;

	int zx = n - mx;


	if (n % 2 == 0) {
		cout << max(cur - zx, 0LL) << endl;
	}
	else {
		cout << max(cur - zx, 1LL) << endl;
	}





}
signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int  t = 1;
	cin >> t;
	while (t--) solve();
	return 0;
}