B. Array Cloning Technique

Problem - 1665B - Codeforces

You are given an array aa of nn integers. Initially there is only one copy of the given array.

You can do operations of two types:

1. Choose any array and clone it. After that there is one more copy of the chosen array.
2. Swap two elements from any two copies (maybe in the same copy) on any positions.

You need to find the minimal number of operations needed to obtain a copy where all elements are equal.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer nn (1≤n≤1051≤n≤105) — the length of the array aa.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109) — the elements of the array aa.

It is guaranteed that the sum of nn over all test cases does not exceed 105105.

Output

For each test case output a single integer — the minimal number of operations needed to create at least one copy where all elements are equal.

Example

input

Copy

6
1
1789
6
0 1 3 3 7 0
2
-1000000000 1000000000
4
4 3 2 1
5
2 5 7 6 3
7
1 1 1 1 1 1 1

output

Copy

0
6
2
5
7
0

Note

In the first test case all elements in the array are already equal, that's why the answer is 00.

In the second test case it is possible to create a copy of the given array. After that there will be two identical arrays:

[ 0 1 3 3 7 0 ][ 0 1 3 3 7 0 ] and [ 0 1 3 3 7 0 ][ 0 1 3 3 7 0 ]

After that we can swap elements in a way so all zeroes are in one array:

[ 0 0– 0– 3 7 0 ][ 0 0_ 0_ 3 7 0 ] and [ 1– 1 3 3 7 3– ][ 1_ 1 3 3 7 3_ ]

Now let's create a copy of the first array:

[ 0 0 0 3 7 0 ][ 0 0 0 3 7 0 ], [ 0 0 0 3 7 0 ][ 0 0 0 3 7 0 ] and [ 1 1 3 3 7 3 ][ 1 1 3 3 7 3 ]

Let's swap elements in the first two copies:

[ 0 0 0 0– 0– 0 ][ 0 0 0 0_ 0_ 0 ], [ 3– 7– 0 3 7 0 ][ 3_ 7_ 0 3 7 0 ] and [ 1 1 3 3 7 3 ][ 1 1 3 3 7 3 ].

Finally, we made a copy where all elements are equal and made 66 operations.

It can be proven that no fewer operations are enough.

#include
#include
#include
using namespace std;
const int N=2e5+10;
typedef long long ll ;
ll a[N];
int t,n;
mapint>mp;
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n;
        int maxx=0;
        ll maxv=0;
        mp.clear();
      fill(a,a+n,0);//用memset会超时
        for(int i=0;i
        {
            cin>>a[i];
            mp[a[i]]++;
            if(mp[a[i]]>maxx)
            {
                maxx=mp[a[i]];
                maxv=a[i];
            }
        }
        if(mp[a[0]]==n)cout<<0<
        else
        {
            int count=0;
            while(mp[maxv]
            {
                count++;//扩展
                if(n-mp[maxv]>=mp[maxv])//供不应求
                {
                    count+=mp[maxv];
                    if(n-mp[maxv]==mp[maxv])break;
                }
                else//通货膨胀，只要少数几个
                {
                    count+=(n-mp[maxv]);
                    break;
                }
                 mp[maxv]+=mp[maxv];//扩展之后的
            }
            cout<
        }
    }
}