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 PYTHON):

Solution:
q = int(input())
for z in range(q):
    n = int(input())
    candies = [int(x) for x in raw_input().split()]

    typeToCount = dict()
    for c in candies:
        typeToCount[c] = 1 if (c not in typeToCount) else typeToCount[c]+1
    # numberOfTypes = 0
    countToTypes = dict()
    
    s = 0
    for t in typeToCount:
        newCount = typeToCount[t]
        while newCount != 0:
            if newCount in countToTypes:
                newCount = countToTypes[newCount][0]
            else:
                if newCount-1 in countToTypes and newCount+1 in countToTypes:
                    countToTypes[newCount] = countToTypes[newCount-1]
                    countToTypes[newCount][1] = countToTypes[newCount+1][1]
                    countToTypes[countToTypes[newCount][1]][0] = countToTypes[newCount][0]
                elif newCount-1 in countToTypes:
                    countToTypes[newCount] = countToTypes[newCount-1]
                    countToTypes[newCount][1] = newCount
                elif newCount+1 in countToTypes:
                    countToTypes[newCount] = countToTypes[newCount+1]
                    countToTypes[newCount][0] = newCount-1 
                else:
                    countToTypes[newCount] = [newCount-1, newCount]

                # if newCount+1 in countToTypes:
                #     countToTypes[newCount] = countToTypes[newCount+1]
                #     countToTypes[newCount][0] = newCount - 1
                # else:
                #     if newCount-1 in countToTypes:
                #         countToTypes[newCount] = countToTypes[newCount-1]
                #     else:
                #         countToTypes[newCount] = [newCount-1]

                # if newCount-1 not in countToTypes:
                #     countToTypes[newCount] = [newCount-1]
                # else:
                #     countToTypes[newCount] = countToTypes[newCount-1]
                # if newCount+1 in countToTypes:
                    
                s += newCount
                break 
    print s
        # if typeToCount[t] in countToTypes:
             
    #srted = sorted(typeToCount.items(), key=lambda item:item[1], reverse=True)

    # runningSum = 0
    # prevCount = -1

    
    # for i in range(len(srted)):
    #     res = srted[i][1]
    #     if srted[i][1] >= prevCount and prevCount >= 0:
    #         res = prevCount - 1
    #     # print res
    #     if res < 0:
    #         break
    #     prevCount = res
    #     runningSum += res
    # # print srted
    # print runningSum
        
    # for k in typeToCount:
    #     numberOfTypes+=1
    #     if typeToCount[k] in countToTypes:
    #         countToTypes[typeToCount[k]].append(k)
    #     else:
    #         countToTypes[typeToCount[k]] = [k]