Problem: This problem is actually a subproblem of problem G from the same contest.

There are n candies in a candy box. The type of the i-th candy is a_i (1 ≤ a_i ≤ n).

You have to prepare a gift using some of these candies with the following restriction: the numbers of candies of each type presented in a gift should be all distinct (i. e. for example, a gift having two candies of type 1 and two candies of type 2 is bad). 

It is possible that multiple types of candies are completely absent from the gift. It is also possible that not all candies of some types will be taken to a gift.

Your task is to find out the maximum possible size of the single gift you can prepare using the candies you have.

You have to answer q independent queries.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries. Each query is represented by two lines.

The first line of each query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of candies.

The second line of each query contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n), where a_i is the type of the i-th candy in the box.

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

Output

For each query print one integer — the maximum possible size of the single gift you can compose using candies you got in this query with the restriction described in the problem statement.

Example

Input


3
8
1 4 8 4 5 6 3 8
16
2 1 3 3 4 3 4 4 1 3 2 2 2 4 1 1
9
2 2 4 4 4 7 7 7 7


Output


3
10
9

Note

In the first query, you can prepare a gift with two candies of type 8 and one candy of type 5, totalling to 3 candies.

Note that this is not the only possible solution — taking two candies of type 4 and one candy of type 6 is also valid.

Solution (in PYTHON3):

Solution:
#!/usr/bin/env python
import os
import operator
import sys
from _collections import defaultdict
import bisect
from io import BytesIO, IOBase

def inar():
    return [int(k) for k in input().split()]
def power(x, p,m):
    res = 1
    while p:
        if p & 1:
            res = (res * x) % m
        x = (x * x) % m
        p >>= 1
    return res
visited=[]
level=[]
adj=defaultdict(list)
def bfs(node):
    global visited, level, adj
    qu=[]
    visited[node]=1
    qu.append(node)
    while qu:
        el=qu.pop(0)
        for child in adj[el]:
            if not visited[child]:
                level[child]=level[el]+1
                visited[child]=1
                qu.append(child)


def main():
    #mod=1000000007
    # global visited, level, adj
    # #for _ in range(int(input())):
    # #n=int(input())
    # n,m,k=map(int,input().split())
    # arr=inar()
    # for i in range(m):
    #     u,v=map(int,input().split())
    #     adj[u].append(v)
    #     adj[v].append(u)
    # level=[0]*(n+1)
    # visited=[0]*(n+1)
    # bfs(1)
    # print(level[n])
    for _ in range(int(input())):
        n=int(input())
        arr=inar()
        freq=defaultdict(int)
        for i in range(n):
            freq[arr[i]]+=1
        freq=dict(sorted(freq.items(),key=operator.itemgetter(1)))
        #print(freq)
        ans=0
        c=0
        ls=list(freq.values())
        ls.sort(reverse=True)
        #print(ls)
        for i in range(len(ls)):
            if i==0:
                ans=ls[i]
                c=ls[i]
                continue
            if c==0:
                break
            if c-1<=ls[i]:
                ans+=(c-1)
                c=c-1
            else:
                ans+=ls[i]
                c=ls[i]
        print(ans)



    # string=input()
    # n=len(string)
    # ans=[]
    # for i in range(26):
    #     lis=[0]*26
    #     ans.append(lis)
    # temp=[0]*26
    # result=0
    # for i in range(0,n):
    #     temp[ord(string[i])-97]+=1
    # for i in range(26):
    #     result=max(result,temp[i])
    # temp=[0]*26
    # temp[ord(string[0])-97]+=1
    #
    # for i in range(1,n):
    #     val=ord(string[i])-97
    #     for j in range(26):
    #         ans[val][j]+=temp[j]
    #     temp[val]+=1
    # # for i in range(26):
    # #     print(*ans[i])
    # for i in range(26):
    #     for j in range(26):
    #         result=max(result,ans[i][j])
    # print(result)




















BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

if __name__ == "__main__":
    main()