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 PYTHON):

Solution:
#   Author: yumtam
#   Created at: 2020-08-29 21:07

from __future__ import division, print_function
_interactive = False

def main():
    for _ in range(int(input())):
        n = int(input())
        ar = [int(c) for c in input()]

        ans = 0
        out = []
        stacks = [[], []]
        for x in ar:
            ox = 1-x
            if stacks[ox]:
                seq = stacks[ox].pop()
                stacks[x].append(seq)
                out.append(seq)
            else:
                ans += 1
                stacks[x].append(ans)
                out.append(ans)

        print(ans)
        print(*out)


# Constants
INF = float('inf')
MOD = 10**9+7

# Python3 equivalent names
import os, sys, itertools
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

# print-flush in interactive problems
if _interactive:
    flush = sys.stdout.flush
    def printf(*args, **kwargs):
        print(*args, **kwargs)
        flush()

# Debug print, only works on local machine
LOCAL = "LOCAL_" in os.environ
debug_print = (print) if LOCAL else (lambda *x, **y: None)

# Fast IO
if (not LOCAL) and (not _interactive):
    from io import BytesIO
    from atexit import register
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
    sys.stdout = BytesIO()
    register(lambda: os.write(1, sys.stdout.getvalue()))
    input = lambda: sys.stdin.readline().rstrip('\r\n')

# Some utility functions(Input, N-dimensional lists, ...)
def input_as_list():
    return [int(x) for x in input().split()]

def input_with_offset(o):
    return [int(x)+o for x in input().split()]

def input_as_matrix(n, m):
    return [input_as_list() for _ in range(n)]

def array_of(f, *dim):
    return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()

# Start of external code templates...
# End of external code templates.

main()