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 CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
const double pi = std::acos(-1);
using namespace std;
const long long int mod = 1000000000 + 7;
long long int power(long long int x, long long int y, long long int md) {
  long long int res = 1;
  x = x % md;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    y = y >> 1;
    x = (x * x) % md;
  }
  return res;
}
long long int m_mul(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  return (a * b + mod) % mod;
}
long long int m_add(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  return (a + b + mod) % mod;
}
long long int spf[(long long int)1e7 + 2] = {0};
void siv() {
  spf[1] = 1;
  for (long long int i = 1; i < 1e7 + 2; i++) spf[i] = i;
  for (long long int i = 2; i < 1e7 + 2; i++) spf[i] = 2, i++;
  for (long long int i = 3; i * i < (long long int)1e7 + 2; i++) {
    if (spf[i] == i)
      for (long long int j = i * i; j < (long long int)1e7 + 2; j = j + i)
        if (spf[j] == j) spf[j] = i;
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long int tc = 1;
  cin >> tc;
  while (tc--) {
    long long int n;
    cin >> n;
    string s;
    cin >> s;
    vector<long long int> v;
    vector<long long int> v0, v1;
    long long int brr[n];
    long long int ans = 0;
    long long int mx = 0;
    long long int gb = 1;
    for (long long int i = 0; i < n; i++) {
      if (s[i] == '0') {
        if (v1.size() == 0) {
          v0.push_back(gb++);
          ans = v0.back();
        } else {
          ans = v1.back();
          v1.pop_back();
          v0.push_back(ans);
        }
      } else {
        if (v0.size() == 0) {
          v1.push_back(gb++);
          ans = v1.back();
        } else {
          ans = v0.back();
          v0.pop_back();
          v1.push_back(ans);
        }
      }
      brr[i] = ans;
      mx = max(brr[i], mx);
    }
    cout << mx << '\n';
    for (long long int i = 0; i < n; i++) cout << brr[i] << " ";
    cout << '\n';
  }
  return 0;
}