This problem is different from the hard version. In this version Ujan makes exactly one exchange. You can hack this problem only if you solve both problems.

After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.

Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation exactly once: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j. Can he succeed?

Note that he has to perform this operation exactly once. He has to perform this operation.

Input

The first line contains a single integer k (1 ≤ k ≤ 10), the number of test cases.

For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 10^4), the length of the strings s and t. 

Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.

Output

For each test case, output "Yes" if Ujan can make the two strings equal and "No" otherwise.

You can print each letter in any case (upper or lower).

Example

Input


4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca


Output


Yes
No
No
No

Note

In the first test case, Ujan can swap characters s_1 and t_4, obtaining the word "house".

In the second test case, it is not possible to make the strings equal using exactly one swap of s_i and t_j.
mod=10**9+7
import sys 
sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
import collections
import math
import heapq
import random
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):                        #for using this uncomment the lines calculating fact and ifact
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x
"""*******************************************************"""
def check(s,a,i,j):
    s[i],a[j]=a[j],s[i]
    if(s==a):
        return 1
    a[j],s[i]=s[i],a[j]
    return 0
def main():
    t=inin()
    ans=[]
    for _ in range(t):
        n=inin()
        s=sin()
        a=sin()
        s=list(s)
        a=list(a)
        an="NO"
        x=[]
        for i in range(n):
            if(a[i]!=s[i]):
                x.append(a[i])
                x.append(s[i])
        if(len(x)==0):
            an="YES"
        elif(len(x)==4):
            if(x[0]==x[2] and x[1]==x[3]):
                an="YES"
        ans.append(an)
    for i in ans:
        print(i)
######## Python 2 and 3 footer by Pajenegod and c1729
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()
