human
PYTHON solution for "Berland Music is a music streaming service built specifically to support Berland local artist. Its developers are currently working on a song recommendation module.

So imagine Monocarp got recommended n songs, numbered from 1 to n. The i-th song had its predicted rating equal to p_i, where 1 ≤ p_i ≤ n and every integer from 1 to n appears exactly once. In other words, p is a permutation.

After listening to each of them, Monocarp pressed either a like or a dislike button. Let his vote sequence be represented with a string s, such that s_i=0 means that he disliked the i-th song, and s_i=1 means that he liked it.

Now the service has to re-evaluate the song ratings in such a way that:

  * the new ratings q_1, q_2, ..., q_n still form a permutation (1 ≤ q_i ≤ n.
each integer from 1 to n appears exactly once).

  * every song that Monocarp liked should have a greater rating than every song that Monocarp disliked (formally, for all i, j such that s_i=1 and s_j=0, q_i>q_j should hold). 


Among all valid permutations q find the one that has the smallest value of ∑_{i=1}^n |p_i-q_i|, where |x| is an absolute value of x.

Print the permutation q_1, q_2, ..., q_n. If there are multiple answers, you can print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.

The first line of each testcase contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of songs.

The second line of each testcase contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n) — the permutation of the predicted ratings.

The third line contains a single string s, consisting of n characters. Each character is either a 0 or a 1. 0 means that Monocarp disliked the song, and 1 means that he liked it.

The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.

Output

For each testcase, print a permutation q — the re-evaluated ratings of the songs. If there are multiple answers such that ∑_{i=1}^n |p_i-q_i| is minimum possible, you can print any of them.

Example

Input

3
2
1 2
10
3
3 1 2
111
8
2...":
assistant
#Here is Python program to find the one that has the smallest value of ∑_{i=1}^n |p_i-q_i|, where |x| is an absolute value of x.
 import sys
testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
interactive = False
if testing:
    cmd = sys.stdout
    from time import time
    start_time = int(round(time() * 1000)) 
    readAll = open(sys.argv[1], 'r').read
    sys.stdout = open(sys.argv[2], 'w')
else:
    readAll = sys.stdin.read

# ############ ---- I/O Functions ---- ############

class InputData:
    def __init__(self):
        self.lines = readAll().split('\n')
        self.n = len(self.lines)
        self.ii = -1
    def input(self):
        self.ii += 1
        assert self.ii < self.n
        return self.lines[self.ii]

flush = sys.stdout.flush
if interactive and not testing:
    input = sys.stdin.readline
else:
    inputData = InputData()
    input = inputData.input

def intin():
    return(int(input()))
def intlin():
    return(list(map(int,input().split())))
def chrin():
    return(list(input()))
def strin():
    return input()
def lout(l, sep="\n", toStr=True):
    print(sep.join(map(str, l) if toStr else l))
def dout(*args, **kargs):
    if not testing: return
    if args: print(args[0] if len(args)==1 else args)
    if kargs: print([(k,v) for k,v in kargs.items()])
def ask(q):
    sys.stdout.write(str(q)+'\n')
    flush()
    return intin()
    
# ############ ---- I/O Functions ---- ############

# from math import ceil
# from collections import defaultdict as ddict, Counter
# from heapq import *
# from Queue import Queue

def main():
    n = intin()
    p = intlin()
    r = strin()
    like = []
    dislike = []
    for i in xrange(n):
        if r[i] == '0':
            like.append((p[i],i))
    for i in xrange(n):
        if r[i] == '1':
            dislike.append((p[i],i))
    like.sort()
    dislike.sort()
    ans = [0]*n
    for i in xrange(len(like)):
        ans[like[i][1]] = i+1
    for j in xrange(len(dislike)):
        ans[dislike[j][1]] = len(like)+j+1
    lout(ans, sep=' ')

anss = []
for _ in xrange(intin()):
    anss.append(main())
    # anss.append("YES" if main() else "NO")

if testing:
    sys.stdout...