Stephen Queen wants to write a story. He is a very unusual writer, he uses only letters 'a', 'b', 'c', 'd' and 'e'!

To compose a story, Stephen wrote out n words consisting of the first 5 lowercase letters of the Latin alphabet. He wants to select the maximum number of words to make an interesting story.

Let a story be a sequence of words that are not necessarily different. A story is called interesting if there exists a letter which occurs among all words of the story more times than all other letters together.

For example, the story consisting of three words "bac", "aaada", "e" is interesting (the letter 'a' occurs 5 times, all other letters occur 4 times in total). But the story consisting of two words "aba", "abcde" is not (no such letter that it occurs more than all other letters in total).

You are given a sequence of n words consisting of letters 'a', 'b', 'c', 'd' and 'e'. Your task is to choose the maximum number of them to make an interesting story. If there's no way to make a non-empty story, output 0.

Input

The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of the words in the sequence. Then n lines follow, each of them contains a word — a non-empty string consisting of lowercase letters of the Latin alphabet. The words in the sequence may be non-distinct (i. e. duplicates are allowed). Only the letters 'a', 'b', 'c', 'd' and 'e' may occur in the words.

It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5; the sum of the lengths of all words over all test cases doesn't exceed 4 ⋅ 10^5.

Output

For each test case, output the maximum number of words that compose an interesting story. Print 0 if there's no way to make a non-empty interesting story.

Example

Input


6
3
bac
aaada
e
3
aba
abcde
aba
2
baba
baba
4
ab
ab
c
bc
5
cbdca
d
a
d
e
3
b
c
ca


Output


3
2
0
2
3
2

Note

In the first test case of the example, all 3 words can be used to make an interesting story. The interesting story is "bac aaada e".

In the second test case of the example, the 1-st and the 3-rd words can be used to make an interesting story. The interesting story is "aba aba". Stephen can't use all three words at the same time.

In the third test case of the example, Stephen can't make a non-empty interesting story. So the answer is 0.

In the fourth test case of the example, Stephen can use the 3-rd and the 4-th words to make an interesting story. The interesting story is "c bc".

'''
The prob i though, had it been that the letter X must appear more times than any other letter
https://codeforces.com/contest/1551/problem/C

Then, each word could be reduced by every abcde until one or more words is lacking.
Some words might become ""

iterate over letters a,b,c,d,e to be considered for our win.
We always want won words, since the contribute to more capacity to other words, naturals dont affect! take all them
for our reduced words, if X occurces more times in a word, then we want it, if it is "" then we want it. For remaining words
we must do knapsack and check the count over all other words so that they are not more than X, we could do for 
size X * 5 - 4 - wwhat we have max array length. Then just pick best over all

something like On2 * something i guess

Turens out, problem was much easier
'''

for _ in range(int(input())):
    n = int(input())
    globans = -1
    # f(w,c) count of c - others in word w
    # 2d arr with w numbererd 0 to n
    f = [[0]*5 for i in range(n)] 
    for i in range(n):
        cnt = [0]*5
        w = input()
        for el in w:
            cnt[ord(el)-97] += 1

        tot = len(w)
        for x in range(5):
            #a,b,c,d,e choice for x
            res = cnt[x] - (tot - cnt[x])
            f[i][x] = res

    for pot in range(5):
        tmp = list()
        for i in range(n):
            tmp.append(f[i][pot])
        
        tmp = sorted(tmp, key=lambda x : -x)
        rol = 0
        idx = 0
        count = 0
        while idx < n:
            rol += tmp[idx]
            if rol <= 0:
                break
            idx += 1
            count += 1
    
        globans = max(count,globans)

    print(globans)
