Problem: The tycoon of a winery empire in Mondstadt, unmatched in every possible way. A thinker in the Knights of Favonius with an exotic appearance.

This time, the brothers are dealing with a strange piece of wood marked with their names. This plank of wood can be represented as a string of n characters. Each character is either a 'D' or a 'K'. You want to make some number of cuts (possibly 0) on this string, partitioning it into several contiguous pieces, each with length at least 1. Both brothers act with dignity, so they want to split the wood as evenly as possible. They want to know the maximum number of pieces you can split the wood into such that the ratios of the number of occurrences of 'D' to the number of occurrences of 'K' in each chunk are the same.

Kaeya, the curious thinker, is interested in the solution for multiple scenarios. He wants to know the answer for every prefix of the given string. Help him to solve this problem!

For a string we define a ratio as a:b where 'D' appears in it a times, and 'K' appears b times. Note that a or b can equal 0, but not both. Ratios a:b and c:d are considered equal if and only if a⋅ d = b⋅ c. 

For example, for the string 'DDD' the ratio will be 3:0, for 'DKD' — 2:1, for 'DKK' — 1:2, and for 'KKKKDD' — 2:4. Note that the ratios of the latter two strings are equal to each other, but they are not equal to the ratios of the first two strings.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.

The first line of each test case contains an integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the length of the wood.

The second line of each test case contains a string s of length n. Every character of s will be either 'D' or 'K'.

It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5.

Output

For each test case, output n space separated integers. The i-th of these numbers should equal the answer for the prefix s_{1},s_{2},...,s_{i}.

Example

Input


5
3
DDK
6
DDDDDD
4
DKDK
1
D
9
DKDKDDDDK


Output


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

Note

For the first test case, there is no way to partition 'D' or 'DDK' into more than one block with equal ratios of numbers of 'D' and 'K', while you can split 'DD' into 'D' and 'D'.

For the second test case, you can split each prefix of length i into i blocks 'D'.

Solution (in CPP):

Solution:
/* Jyothi swaroop reddy*/

#include<bits/stdc++.h>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace chrono;
//using namespace __gnu_pbds;

#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define MOD 1000000007
#define MOD1 998244353
#define INF 1e18
#define endl "\n"
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.141592653589793238462
#define set_bits __builtin_popcountint
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define int long long
#define intl unsigned long long
#define ld long double
//typedef tree < int ,  null_type ,  less<int>  ,  rb_tree_tag ,  tree_order_statistics_node_update> data_set; // find_by_order, order_of_key
/*---------------------------------------------------------------------------------------------------------------------------*/
int gcd(int a, int b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}
int expo(int a, int b, int mod) {int res = 1; while (b > 0) {if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1;} return res;}
void extendgcd(int a, int b, int*v) {if (b == 0) {v[0] = 1; v[1] = 0; v[2] = a; return ;} extendgcd(b, a % b, v); int x = v[1]; v[1] = v[0] - v[1] * (a / b); v[0] = x; return;} //pass an arry of size1 3
int mminv(int a, int b) {int arr[3]; extendgcd(a, b, arr); return arr[0];} //for non prime b
int mminvprime(int a, int b) {return expo(a, b - 2, b);}
bool revsort(int a, int b) {return a > b;}
void swap(int &x, int &y) {int temp = x; x = y; y = temp;}
int combination(int n, int r, int m, int *fact, int *ifact) {int val1 = fact[n]; int val2 = ifact[n - r]; int val3 = ifact[r]; return (((val1 * val2) % m) * val3) % m;}
void google(int t) {cout << "Case #" << t << ": ";}
vector<int> sieve(int n) {int*arr = new int[n + 1](); vector<int> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}
int mod_add(int a, int b, int m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
int mod_mul(int a, int b, int m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
int mod_sub(int a, int b, int m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
int mod_div(int a, int b, int m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;}  //only for prime m
int phin(int n) {int number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (int i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))
/*--------------------------------------------------------------------------------------------------------------------------*/
const int maxN = 200000;
/*
int binarysearch(int st,int e)
{
    int mid=(st+e)/2;
    while(st<=e)
    {
        if(test())
        {
            st=mid+1;
            mid=(st+e)/2;
        }
        else{
            e=mid-1;
            mid=(st+e)/2;
        }
    }
    return mid;
}*/
//vector<vector<int> > g(200001);
//int dp[200001];
//int dp[3001][3001];
map<double,vector<int> > h;
//multiset<int> s;
//queue<int> q;
// less-L to S,greater-S to L
//priority_queue<int> q;
void solve()
{
   int t=1;
   cin>>t;
   while(t--)
   {
      int n;cin>>n;
      string s;cin>>s;
      vector<int> d(n,0);
      vector<int> k(n,0);
      d[0]=(s[0]=='D');
      k[0]=(s[0]=='K');
      cout<<1<<" ";
      for(int i=1;i<n;i++)
      {
          d[i]=d[i-1]+(s[i]=='D');
          k[i]=k[i-1]+(s[i]=='K');
          if(d[i]>=1 &&k[i]>=1 )
          {
              double pk=d[i];
              pk=pk/k[i];
              h[pk].pb(i);
              cout<<h[pk].size()<<" ";
          }
          else if(d[i]>0 && k[i]==0)
          {
              cout<<d[i]<<" ";
          }
          else if(k[i]>0 && d[i]==0)
          {
              cout<<k[i]<<" ";
          }
          else{
            cout<<1<<" ";
          }
      }
      cout<<endl;
      h.clear();
   }
}
int32_t main()
{
    fastio();
    auto start1 = high_resolution_clock::now();
    solve();
    auto stop1 = high_resolution_clock::now();
    auto duration = duration_cast<microseconds>(stop1 - start1);
}