Problem: You are given a binary string s consisting of n zeros and ones.

Your task is to divide the given string into the minimum number of subsequences in such a way that each character of the string belongs to exactly one subsequence and each subsequence looks like "010101 ..." or "101010 ..." (i.e. the subsequence should not contain two adjacent zeros or ones).

Recall that a subsequence is a sequence that can be derived from the given sequence by deleting zero or more elements without changing the order of the remaining elements. For example, subsequences of "1011101" are "0", "1", "11111", "0111", "101", "1001", but not "000", "101010" and "11100".

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s. The second line of the test case contains n characters '0' and '1' — the string s.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer: in the first line print one integer k (1 ≤ k ≤ n) — the minimum number of subsequences you can divide the string s to. In the second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the number of subsequence the i-th character of s belongs to.

If there are several answers, you can print any.

Example

Input


4
4
0011
6
111111
5
10101
8
01010000


Output


2
1 2 2 1 
6
1 2 3 4 5 6 
1
1 1 1 1 1 
4
1 1 1 1 1 2 3 4 

Solution (in PYTHON3):

Solution:
import time,math as mt,bisect,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
    return int(stdin.readline())
def IO(): # to take string input
    return stdin.readline()
def IP(): # to take tuple as input
    return map(int,stdin.readline().split())
def L(): # to take list as input
    return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
    return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
    return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
    return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
    if a==0:
        return b
    elif b==0:
        return a
    if a>b:
        return gcd(a%b,b)
    else:
        return gcd(a,b%a)
def readTree(): # to read tree
    v=int(input())
    adj=[set() for i in range(v+1)]
    for i in range(v-1):
        u1,u2=In()
        adj[u1].add(u2)
        adj[u2].add(u1)
    return adj,v
def bfs(adj,v): # a schema of bfs
    visited=[False]*(v+1)
    q=deque()
    while q:
        pass
def sieve():
    li=[True]*1000001
    li[0],li[1]=False,False
    for i in range(2,len(li),1):
        if li[i]==True:
            for j in range(i*i,len(li),i):
                li[j]=False
    prime=[]
    for i in range(1000001):
        if li[i]==True:
            prime.append(i)
    return prime
def setBit(n):
    count=0
    while n!=0:
        n=n&(n-1)
        count+=1
    return count
mx=10**7
spf=[mx]*(mx+1)
def SPF():
    spf[1]=1
    for i in range(2,mx+1):
        if spf[i]==mx:
            spf[i]=i
            for j in range(i*i,mx+1,i):
                if i<spf[j]:
                    spf[j]=i
    return
#####################################################################################
mod=10**9+7
def solve():
    n=II()
    s=input()
    if n==1:
        print(1)
        print(1)
        return
    sub0,sub1=deque(),deque()
    pos=[-1]*n
    v=1
    for i in range(n):
        if s[i]=='0':
            if sub1:
                x=sub1.pop()
                sub0.append(x)
                pos[i]=x
            else:
                sub0.append(v)
                pos[i]=v
                v+=1
        else:
            if sub0:
                x=sub0.pop()
                sub1.append(x)
                pos[i]=x
            else:
                sub1.append(v)
                pos[i]=v
                v+=1
    print(max(pos))
    print(*pos)

t=II()
for i in range(t):
    solve()

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