human
Write me a Python program to print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value given the following:
Ashish has n elements arranged in a line. 

These elements are represented by two integers a_i — the value of the element and b_i — the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.

He can perform the following operation any number of times: 

  * Select any two elements i and j such that b_i ≠ b_j and swap them. That is, he can only swap two elements of different types in one move. 


Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains one integer n (1 ≤ n ≤ 500) — the size of the arrays.

The second line contains n integers a_i (1 ≤ a_i ≤ 10^5) — the value of the i-th element.

The third line containts n integers b_i (b_i ∈ \{0, 1\}) — the type of the i-th element.

Output

For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.

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

Example

Input

5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1

Output

Yes
Yes
Yes
No
Yes

Note

For the first case: The elements are already in sorted order.

For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.

For the third case: The elements are already in sorted order.

For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i ≠ b_j. The elements cannot be sorted.

For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2.
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python program to print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.

import os
import heapq
import sys
import math
import operator
from collections import defaultdict
from io import BytesIO, IOBase
# def gcd(a,b):
#     if b==0:

#         return a
#     else:
#         return gcd(b,a%b)
def inar():
    return [int(k) for k in input().split()]

# def bubbleSort(arr,b):
#     n = len(arr)
#     for i in range(n):
#         for j in range(0, n - i - 1):
#             if arr[j] > arr[j + 1] and b[j]!=b[j+1]:
#                 arr[j], arr[j + 1] = arr[j + 1], arr[j]
#                 b[j],b[j+1]=b[j+1],b[j]

def main():
    for _ in range(int(input())):
        n=int(input())
        a=inar()
        b=inar()
        lis=a.copy()
        lis.sort()
        z=0
        o=0
        for i in range(n):
            if b[i]==1:
                o+=1
            else:
                z+=1
        if z!=0 and o!=0:
            print("Yes")
        else:
            if lis==a:
                print("Yes")
            else:
                print("No")






BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
           ...