You are given an array of integers a of length n. The elements of the array can be either different or the same. 

Each element of the array is colored either blue or red. There are no unpainted elements in the array. One of the two operations described below can be applied to an array in a single step:

  * either you can select any blue element and decrease its value by 1; 
  * or you can select any red element and increase its value by 1. 



Situations in which there are no elements of some color at all are also possible. For example, if the whole array is colored blue or red, one of the operations becomes unavailable.

Determine whether it is possible to make 0 or more steps such that the resulting array is a permutation of numbers from 1 to n?

In other words, check whether there exists a sequence of steps (possibly empty) such that after applying it, the array a contains in some order all numbers from 1 to n (inclusive), each exactly once.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4) — the number of input data sets in the test.

The description of each set of input data consists of three lines. The first line contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the original array a. The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the array elements themselves.

The third line has length n and consists exclusively of the letters 'B' and/or 'R': ith character is 'B' if a_i is colored blue, and is 'R' if colored red.

It is guaranteed that the sum of n over all input sets does not exceed 2 ⋅ 10^5.

Output

Print t lines, each of which contains the answer to the corresponding test case of the input. Print YES as an answer if the corresponding array can be transformed into a permutation, and NO otherwise.

You can print the answer in any case (for example, the strings yEs, yes, Yes, and YES will be recognized as a positive answer).

Example

Input


8
4
1 2 5 2
BRBR
2
1 1
BB
5
3 1 4 2 5
RBRRB
5
3 1 3 1 3
RBRRB
5
5 1 5 1 5
RBRRB
4
2 2 2 2
BRBR
2
1 -2
BR
4
-2 -1 4 0
RRRR


Output


YES
NO
YES
YES
NO
YES
YES
YES

Note

In the first test case of the example, the following sequence of moves can be performed:

  * choose i=3, element a_3=5 is blue, so we decrease it, we get a=[1,2,4,2]; 
  * choose i=2, element a_2=2 is red, so we increase it, we get a=[1,3,4,2]; 
  * choose i=3, element a_3=4 is blue, so we decrease it, we get a=[1,3,3,2]; 
  * choose i=2, element a_2=2 is red, so we increase it, we get a=[1,4,3,2]. 



We got that a is a permutation. Hence the answer is YES.
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######  ######  #########  #######  ######               #############   ################  ############
######         ##########  #######  ######  ###########  #############   ################  ############
######  #####  ##########  #######  ######  ###########  #############   ################  ############
######  ######  ##########  #####  #######  ###########  #############   ################  ############
####     ######     ######        ######    ########      #######            ##########     ###########
#######################################################################################################
import os
import sys
from io import BytesIO, IOBase
#-----------------------IMPORTANT  FUNCTIONS ---------------------------------#
from math import cos, pi, ceil, pi, tan, floor, gcd, sqrt, log2, factorial, log10
from collections import Counter
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase
from statistics import mean, median, mode
from bisect import *

alpha = ascii_lowercase
beta = ascii_uppercase


#-----------------------NEXT POWER OF 2---------------------------------------#
def nextPowerOf2(n):
    p = 1
    if (n and not (n & (n - 1))):
        return n
    while (p < n):
        p <<= 1
    return p


#-----------------------COMPOSITE  CHECK---------------------------------------#
def isComposite(n):
    if (n <= 1):
        return False
    if (n <= 3):
        return False
    if (n % 2 == 0 or n % 3 == 0):
        return True
    i = 5
    while (i * i <= n):
        if (n % i == 0 or n % (i + 2) == 0):
            return True
        i = i + 6
    return False


#-----------------------GET DIVISORS 1 TO <=N   ------------------------------#
def get_Divisors(n):
    arr = []
    i = 1
    while i <= sqrt(n):
        if (n % i == 0):
            if (n / i == i):
                arr.append(i)
            else:
                arr.append(i)
                arr.append(n // i)
        i = i + 1

    return arr


#----------------------CHECK    PRIMES-----------------------------------#
def isprime(n):
    if n == 1:
        return 0
    i = 2
    while (i * i <= n):
        if (n % i == 0):
            return 0
        i += 1
    return 1


#-----------------------PRE COMPUTED  PRIMES-----------------------------------#
def Sieve(n):
    prime = [True for i in range(n + 1)]
    prime[0] = False
    prime[1] = False
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
    return prime


#-----------------------LARGER VALUE OF  X,Y,10**9+7 --------------------------#
def power(x, y, p):
    res = 1
    x = x % p
    if (x == 0):
        return 0
    while (y > 0):
        if ((y & 1) == 1):
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res


#-----------------------       CREATE MATRIX         --------------------------#
def create_matrix(n, m):
    mat = [[] for i in range(n)]
    for i in range(n):
        for j in range(m):
            mat[i].append(0)
    return mat


#By Default Counter stores the value in decending orderCounter({4: 2, 8: 2, 2: 1, 9: 1}
#s=sorted(s.items(),key=lambda k:(k[0],k[1])) ---> [(2, 1), (4, 2), (8, 2), (9, 1)]
#sort 2nd value  from bada se chhota  then(4,2)(8,2)->>(8,2)(4,2) s=sorted(s.items(),key=lambda k:(-k[1],-k[0]))  [(8, 2), (4, 2), (9, 1), (2, 1)]
#for list pair in a list    s.sort(key=lambda k:(-k[1],-k[0])) [[3, 4], [8, 2], [4, 2]]
# from functools import cmp_to_key
#how to use compare ----ss = sorted(s, key=cmp_to_key(compare))
#-----------------------*****CODE STARTS  HERE*****--------------------------#

# from queue import Queue
# from random import *
def main():
    for ttt in range(int(input())):
        n=int(input())
        arr=list(map(int,input().split()))
        s=input()
        res=[]
        ans=[]
        for i in range(n):
            if s[i]=='B':
                res.append([arr[i],'B'])
            else:     
                ans.append([arr[i],'R'])
        ans.sort()
        res.sort()
        ss=res+ans 
        for i in range(n):
            if ss[i][1]=='B' and ss[i][0]>=(i+1):
                ss[i][0]=(i+1)
            elif ss[i][1]=='R' and ss[i][0]<=(i+1):
                ss[i][0]=(i+1)
        for i in range(n):
            if ss[i][0]!=(i+1):
                print("NO")
                break 
        else:
            print("YES")
        
# Region Fast_______io
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:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        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")
if __name__ == "__main__":
    main()