Let's denote the f(x) function for a string x as the number of distinct characters that the string contains. For example f(abc)=3, f(bbbbb)=1, and f(babacaba)=3.

Given a string s, split it into two non-empty strings a and b such that f(a)+f(b) is the maximum possible. In other words, find the maximum possible value of f(a)+f(b) such that a+b=s

?(the concatenation of string a and string b is equal to string s).

Input

The input consists of multiple test cases. The first line contains an integer t (1≤t≤104) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (2≤n≤2?105) — the length of the string s.

The second line contains the string s, consisting of lowercase English letters.

It is guaranteed that the sum of n over all test cases does not exceed 2?105.

Output

For each test case, output a single integer? — the maximum possible value of f(a)+f(b) such that a+b=s.

Example

input

5

2

aa

7

abcabcd

5

aaaaa

10

paiumoment

4

aazz

output

2

7

2

10

3

Note

For the first test case, there is only one valid way to split aa into two non-empty strings a and a, and f(a)+f(a)=1+1=2.

For the second test case, by splitting abcabcd into abc and abcd we can get the answer of f(abc)+f(abcd)=3+4=7 which is maximum possible.

For the third test case, it doesn't matter how we split the string, the answer will always be 2.

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維護(hù)2個(gè)map先匯總當(dāng)前所有的字符串的distinct的數(shù)量，然后依次從前面開始遍歷，每次放進(jìn)一個(gè)新的map中，同時(shí)將之前舊的map的值-1，如果值等于0 了，那么就刪除這個(gè)key，

然后用max保存每次2個(gè)map的size。最后輸出即可，下面是代碼；