# SOLUTION:- CLICK HERE

A (11-indexed) binary string SS of length NN is called a xor palindrome if the value of SiS(N+1i)Si⊕S(N+1−i) is the same for all 1iN1≤i≤N.

For example, 0011111111 and 01010101 are xor palindromes, while 11101110 and 110101110101 are not.

You are given a binary string SS of length NN. Determine if it is possible to rearrange it to form a xor palindrome or not.

### Input Format

• The first line of input contains a single integer TT — the number of test cases. The description of TT test cases follows.
• The first line of each test case contains an integer NN — the length of the binary string SS.
• The second line of each test case contains the binary string SS containing 00s and 11s only.

### Output Format

For each test case, output YESYES if it is possible to rearrange SS to convert it into a xor palindrome. Otherwise output NONO.

You may print each character of YESYES and NONO in uppercase or lowercase (for example, yesyesyEsyEsYesYes will be considered identical).

### Constraints

• 1T10001≤T≤1000
• 1N1051≤N≤105
• SS is a binary string, i.e, contains only the characters 00 and 11
• It is guaranteed that the sum of NN over all test cases does not exceed 21052⋅105.

### Subtasks

Subtask #1 (100 points): Original constraints

### Sample Input 1

4
2
00
4
0011
3
001
4
0001


### Sample Output 1

YES
YES
YES
NO


### Explanation

Test case 110000 is already a xor palindrome. [The value of SiS(N+1i)Si⊕S(N+1−i) is 00 for all 1iN1≤i≤N.]

Test case 2200110011 is already a xor palindrome. [The value of SiS(N+1i)Si⊕S(N+1−i) is 11 for all 1iN1≤i≤N.]

Test case 33001001 can be rearranged to form 010010 which is a xor palindrome. [The value of SiS(N+1i)Si⊕S(N+1−i) is 00 for all 1iN1≤i≤N.]

Test case 44: It can be proved that 00010001 can not be rearranged to form a xor palindrome.