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 sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
# import numpy as np


mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]

try:
    # sys.setrecursionlimit(int(pow(10,6)))
    sys.stdin = open("input.txt", "r")
    sys.stdout = open("../output.txt", "w")
except:
    pass


# @lru_cache(None)
class solve():
    def __init__(self):
        self.d={}
        self.ptr=0
        self.one=[]
        self.zero=[]
        self.ans={}
    def x(self,idx):
        if s[idx]=="1":
            if not self.zero:
                self.one.append(self.ptr+1)
                self.ptr+=1
                temp=self.d.get(self.ptr,[])
                temp.append(idx)
                self.d[self.ptr]=temp
                self.ans[idx]=self.ptr
            else:
                a=self.zero[0]
                del(self.zero[0])
                self.one.append(a)
                self.d[a].append(idx)
                self.ans[idx]=a
        elif s[idx]=="0":
            if not self.one:
                self.zero.append(self.ptr+1)
                self.ptr+=1
                temp=self.d.get(self.ptr,[])
                temp.append(idx)
                self.d[self.ptr]=temp
                self.ans[idx]=self.ptr
            else:
                a=self.one[0]
                del(self.one[0])
                self.zero.append(a)
                self.d[a].append(idx)
                self.ans[idx]=a
    def printsol(self):
        print(self.ptr)
        for i in range(n):
            print(self.ans[i],end=" ")
        print()

                

for _ in range(L()[0]):
    n=L()[0]
    s=input()
    obj=solve()

    for i in range(n):
        obj.x(i)
    obj.printsol()