You are given two strings $s$ and $t$ both of length $n$ and both consisting of lowercase Latin letters.
In one move, you can choose any length $len$ from $1$ to $n$ and perform the following operation:   Choose any contiguous substring of the string $s$ of length $len$ and reverse it;  at the same time choose any contiguous substring of the string $t$ of length $len$ and reverse it as well. 
Note that during one move you reverse exactly one substring of the string $s$ and exactly one substring of the string $t$.
Also note that borders of substrings you reverse in $s$ and in $t$ can be different, the only restriction is that you reverse the substrings of equal length. For example, if $len=3$ and $n=5$, you can reverse $s[1 \dots 3]$ and $t[3 \dots 5]$, $s[2 \dots 4]$ and $t[2 \dots 4]$, but not $s[1 \dots 3]$ and $t[1 \dots 2]$.
Your task is to say if it is possible to make strings $s$ and $t$ equal after some (possibly, empty) sequence of moves.
You have to answer $q$ independent test cases.
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of test cases. Then $q$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $s$ and $t$.
The second line of the test case contains one string $s$ consisting of $n$ lowercase Latin letters.
The third line of the test case contains one string $t$ consisting of $n$ lowercase Latin letters.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).
-----Output-----
For each test case, print the answer on it — "YES" (without quotes) if it is possible to make strings $s$ and $t$ equal after some (possibly, empty) sequence of moves and "NO" otherwise.
-----Example-----
Input
4
4
abcd
abdc
5
ababa
baaba
4
asdf
asdg
4
abcd
badc
Output
NO
YES
NO
YES
def cal(s):
    res = 0
    for i in range(len(s)):
        for j in range(i+1,len(s)):
            if s[i]>s[j]:
                res ^= 1
    return res

def solve():
    int(input())
    s = input()
    t = input()
    if(sorted(s)!=sorted(t)):
        print('NO')
        return
    for i in range(26):
        if s.count(chr(ord('a')+i))>1:
            print('YES')
            return
    print(['NO','YES'][cal(s)==cal(t)])

for i in range(int(input())):
    solve()

def cal(s):
    res = 0
    for i in range(len(s)):
        for j in range(i+1,len(s)):
            if s[i]>s[j]:
                res ^= 1
    return res

def solve():
    int(input())
    s = input()
    t = input()
    if(sorted(s)!=sorted(t)):
        print('NO')
        return
    for i in range(26):
        if s.count(chr(ord('a')+i))>1:
            print('YES')
            return
    print(['NO','YES'][cal(s)==cal(t)])

for i in range(int(input())):
    solve()
