#pragma GCC target("avx2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
#define DEBUG
#ifdef DEBUG
template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
    os << '(' << p.first << ',' << p.second << ')';
    return os;
}
template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
    os << '{';
    for(int i = 0; i < (int)v.size(); i++) {
        if(i) { os << ','; }
        os << v[i];
    }
    os << '}';
    return os;
}
void debugg() { cerr << endl; }
template <class T, class... Args>
void debugg(const T &x, const Args &... args) {
    cerr << " " << x;
    debugg(args...);
}
#define debug(...)                                                             \
    cerr << __LINE__ << " [" << #__VA_ARGS__ << "]: ", debugg(__VA_ARGS__)
#define dump(x) cerr << __LINE__ << " " << #x << " = " << (x) << endl
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef pair<ll,ll> P;
typedef pair<int,int> pii;
typedef vector<P> vpl;
typedef tuple<ll,ll,ll> tapu;
#define rep(i,n) for(int i=0; i<(n); i++)
#define REP(i,a,b) for(int i=(a); i<(b); i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
const int inf = 1<<30;
const ll linf = 1LL<<62;
const int MAX = 4020000;
int dy[8] = {0,1,0,-1,1,-1,-1,1};
int dx[8] = {-1,0,1,0,1,-1,1,-1};
const double pi = acos(-1);
const double eps = 1e-7;
template<typename T1,typename T2> inline bool chmin(T1 &a,T2 b){
	if(a>b){
		a = b; return true;
	}
	else return false;
}
template<typename T1,typename T2> inline bool chmax(T1 &a,T2 b){
	if(a<b){
		a = b; return true;
	}
	else return false;
}
template<typename T> inline void print(T &a){
    int sz = a.size();
    for(auto itr = a.begin(); itr != a.end(); itr++){
		cout << *itr;
        sz--;
        if(sz) cout << " ";
	}
    cout << "\n";
}
template<typename T1,typename T2> inline void print2(T1 a, T2 b){
	cout << a << " " << b << "\n";
}
template<typename T1,typename T2,typename T3> inline void print3(T1 a, T2 b, T3 c){
	cout << a << " " << b << " " << c << "\n";
}
void mark() {cout << "#" << "\n";}
ll pcount(ll x) {return __builtin_popcountll(x);}
const int mod = 1e9 + 7;
//const int mod = 998244353;


void solve(){
    int n; cin >> n;
    vl cnt(n+2,0);
    map<int,int> mp;
    rep(i,n){
        int a; cin >> a;
        mp[a]++;
    }
    for(auto x : mp) cnt[x.second]++;
    ll ans = 0, res = 0, m = 0;
    for(ll i=2; i<=n; i++){
        res += cnt[i] * (i-1);
        m += cnt[i];
    }
    ans = res;
    for(ll i=1; i<=n; i++){
        res += cnt[i] * i;
        res -= m;
        m -= cnt[i+1];
        chmin(ans, res);
        //debug(res);
    }
    cout << ans << "\n";
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t = 1;
    cin >> t;
    while(t--){
        solve();
    }
}
### Prompt

Create a solution in cpp for the following problem:

Polycarp was gifted an array a of length n. Polycarp considers an array beautiful if there exists a number C, such that each number in the array occurs either zero or C times. Polycarp wants to remove some elements from the array a to make it beautiful.

For example, if n=6 and a = [1, 3, 2, 1, 4, 2], then the following options are possible to make the array a array beautiful: 

  * Polycarp removes elements at positions 2 and 5, array a becomes equal to [1, 2, 1, 2]; 
  * Polycarp removes elements at positions 1 and 6, array a becomes equal to [3, 2, 1, 4]; 
  * Polycarp removes elements at positions 1, 2 and 6, array a becomes equal to [2, 1, 4]; 



Help Polycarp determine the minimum number of elements to remove from the array a to make it beautiful.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case consists of one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the 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) — array a.

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

Output

For each test case, output one integer — the minimum number of elements that Polycarp has to remove from the array a to make it beautiful.

Example

Input


3
6
1 3 2 1 4 2
4
100 100 4 100
8
1 2 3 3 3 2 6 6


Output


2
1
2

### Response

```cpp
#pragma GCC target("avx2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
#define DEBUG
#ifdef DEBUG
template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
    os << '(' << p.first << ',' << p.second << ')';
    return os;
}
template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
    os << '{';
    for(int i = 0; i < (int)v.size(); i++) {
        if(i) { os << ','; }
        os << v[i];
    }
    os << '}';
    return os;
}
void debugg() { cerr << endl; }
template <class T, class... Args>
void debugg(const T &x, const Args &... args) {
    cerr << " " << x;
    debugg(args...);
}
#define debug(...)                                                             \
    cerr << __LINE__ << " [" << #__VA_ARGS__ << "]: ", debugg(__VA_ARGS__)
#define dump(x) cerr << __LINE__ << " " << #x << " = " << (x) << endl
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef pair<ll,ll> P;
typedef pair<int,int> pii;
typedef vector<P> vpl;
typedef tuple<ll,ll,ll> tapu;
#define rep(i,n) for(int i=0; i<(n); i++)
#define REP(i,a,b) for(int i=(a); i<(b); i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
const int inf = 1<<30;
const ll linf = 1LL<<62;
const int MAX = 4020000;
int dy[8] = {0,1,0,-1,1,-1,-1,1};
int dx[8] = {-1,0,1,0,1,-1,1,-1};
const double pi = acos(-1);
const double eps = 1e-7;
template<typename T1,typename T2> inline bool chmin(T1 &a,T2 b){
	if(a>b){
		a = b; return true;
	}
	else return false;
}
template<typename T1,typename T2> inline bool chmax(T1 &a,T2 b){
	if(a<b){
		a = b; return true;
	}
	else return false;
}
template<typename T> inline void print(T &a){
    int sz = a.size();
    for(auto itr = a.begin(); itr != a.end(); itr++){
		cout << *itr;
        sz--;
        if(sz) cout << " ";
	}
    cout << "\n";
}
template<typename T1,typename T2> inline void print2(T1 a, T2 b){
	cout << a << " " << b << "\n";
}
template<typename T1,typename T2,typename T3> inline void print3(T1 a, T2 b, T3 c){
	cout << a << " " << b << " " << c << "\n";
}
void mark() {cout << "#" << "\n";}
ll pcount(ll x) {return __builtin_popcountll(x);}
const int mod = 1e9 + 7;
//const int mod = 998244353;


void solve(){
    int n; cin >> n;
    vl cnt(n+2,0);
    map<int,int> mp;
    rep(i,n){
        int a; cin >> a;
        mp[a]++;
    }
    for(auto x : mp) cnt[x.second]++;
    ll ans = 0, res = 0, m = 0;
    for(ll i=2; i<=n; i++){
        res += cnt[i] * (i-1);
        m += cnt[i];
    }
    ans = res;
    for(ll i=1; i<=n; i++){
        res += cnt[i] * i;
        res -= m;
        m -= cnt[i+1];
        chmin(ans, res);
        //debug(res);
    }
    cout << ans << "\n";
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t = 1;
    cin >> t;
    while(t--){
        solve();
    }
}
```