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 JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.BigInteger;

/**
 __                  __
 ( _)                ( _)
 / / \\              / /\_\_
 / /   \\            / / | \ \
 / /     \\          / /  |\ \ \
 /  /   ,  \ ,       / /   /|  \ \
 /  /    |\_ /|      / /   / \   \_\
 /  /  |\/ _ '_| \   / /   /   \    \\
 |  /   |/  0 \0\    / |    |    \    \\
 |    |\|      \_\_ /  /    |     \    \\
 |  | |/    \.\ o\o)  /      \     |    \\
 \    |     /\\`v-v  /        |    |     \\
 | \/    /_| \\_|  /         |    | \    \\
 | |    /__/_ `-` /   _____  |    |  \    \\
 \|    [__]  \_/  |_________  \   |   \    ()
 /    [___] (    \         \  |\ |   |   //
 |    [___]                  |\| \|   /  |/
 /|    [____]                  \  |/\ / / ||
 (  \   [____ /     ) _\      \  \    \| | ||
 \  \  [_____|    / /     __/    \   / / //
 |   \ [_____/   / /        \    |   \/ //
 |   /  '----|   /=\____   _/    |   / //
 __ /  /        |  /   ___/  _/\    \  | ||
 (/-(/-\)       /   \  (/\/\)/  |    /  | /
 (/\/\)           /   /   //
 _________/   /    /
 \____________/    (


 @author NTUDragons-Reborn
 */

public class Solution{
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    // main solver
    static class Task{

        double eps= 0.00000001;
        static final int MAXN = 100005;
        static final int MOD= 1000000007;

        // stores smallest prime factor for every number
        static int spf[] = new int[MAXN];

        Map<Integer,Set<Integer>> dp= new HashMap<>();

        // Calculating SPF (Smallest Prime Factor) for every
        // number till MAXN.
        // Time Complexity : O(nloglogn)
        public void sieve()
        {
            spf[1] = 1;
            for (int i=2; i<MAXN; i++)

                // marking smallest prime factor for every
                // number to be itself.
                spf[i] = i;

            // separately marking spf for every even
            // number as 2
            for (int i=4; i<MAXN; i+=2)
                spf[i] = 2;

            for (int i=3; i*i<MAXN; i++)
            {
                // checking if i is prime
                if (spf[i] == i)
                {
                    // marking SPF for all numbers divisible by i
                    for (int j=i*i; j<MAXN; j+=i)

                        // marking spf[j] if it is not
                        // previously marked
                        if (spf[j]==j)
                            spf[j] = i;
                }
            }
        }

        // A O(log n) function returning primefactorization
        // by dividing by smallest prime factor at every step
        public Set<Integer> getFactorization(int x)
        {
            if(dp.containsKey(x)) return dp.get(x);
            Set<Integer> ret = new HashSet<>();
            while (x != 1)
            {
                if(spf[x]!=2) ret.add(spf[x]);
                x = x / spf[x];
            }
            dp.put(x,ret);
            return ret;
        }
        // function to find first index >= x
        public int lowerIndex(List<Integer> arr, int n, int x)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) >= x)
                    h = mid - 1;
                else
                    l = mid + 1;
            }
            return l;
        }

        // function to find last index <= y
        public int upperIndex(List<Integer> arr, int n, int y)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) <= y)
                    l = mid + 1;
                else
                    h = mid - 1;
            }
            return h;
        }

        // function to count elements within given range
        public int countInRange(List<Integer> arr, int n, int x, int y)
        {
            // initialize result
            int count = 0;
            count = upperIndex(arr, n, y) -
                    lowerIndex(arr, n, x) + 1;
            return count;
        }

        public int _gcd(int a,  int  b)
        {

            if(b == 0) {
                return a;
            }
            else {
                return _gcd(b, a % b);
            }
        }

        public int add(int a, int b){
            a+=b;
            if(a>=MOD) a-=MOD;
            else if(a<0) a+=MOD;
            return a;
        }

        public int mul(int a, int b){
            long res= (long)a*(long)b;
            return (int)(res%MOD);
        }

        public int power(int a, int b) {
            int ans=1;
            while(b>0){
                if((b&1)!=0) ans= mul(ans,a);
                b>>=1;
                a= mul(a,a);
            }
            return ans;
        }

        int[] fact= new int[MAXN];
        int[] inv= new int[MAXN];

        public int Ckn(int n, int k){
            if(k<0 || n<0) return 0;
            return mul(mul(fact[n],inv[k]),inv[n-k]);
        }

        public int inverse(int a){
            return power(a,MOD-2);
        }

        public void preprocess() {
            fact[0]=1;
            for(int i=1;i<MAXN;i++) fact[i]= mul(fact[i-1],i);
            inv[MAXN-1]= inverse(fact[MAXN-1]);
            for(int i=MAXN-2;i>=0;i--){
                inv[i]= mul(inv[i+1],i+1);
            }
        }
        
        InputReader in; PrintWriter out;
        
        public void solve(InputReader in, PrintWriter out) {
            this.in=in; this.out=out;
            int t= in.nextInt();
            while(t-->0){
                int n= in.nextInt();
                char[] s= in.nextStringCharArr();
                int[] preK= new int[n+1];
                int[] preD= new int[n+1];
                for(int i=0;i<n;i++){
                    preK[i+1]= preK[i]+(s[i]=='K'?1:0);
                    preD[i+1]= preD[i]+(s[i]=='D'?1:0);
                }
                int[] res= new int[n+1];
                Map<Double,Integer> cnt= new HashMap<>();
                for(int i=1;i<=n;i++){
                    double hash=-1;
                    if(preK[i]==0 || preD[i]==0) hash=0;
                    else hash= (double)preD[i]/preK[i];
                    cnt.put(hash, cnt.getOrDefault(hash,0)+1);
                    res[i]=cnt.get(hash);
                }
                for(int i=1;i<=n;i++) out.print(res[i]+" ");
                out.println();
            }
        }

    }

    static class Venice{
        public Map<Long,Long> m= new HashMap<>();
        public long base=0;
        public long totalValue=0;
        private int M= 1000000007;

        private long addMod(long a, long b){
            a+=b;
            if(a>=M) a-=M;
            return a;
        }

        public void reset(){
            m= new HashMap<>();
            base=0;
            totalValue=0;
        }
        public void update(long add){
            base= base+ add;
        }
        public void add(long key, long val){
            long newKey= key-base;
            m.put(newKey, addMod(m.getOrDefault(newKey,(long)0),val));
        }
    }


    static class Tuple implements Comparable<Tuple>{
        int x, y, z;
        public Tuple(int x, int y, int z){
            this.x= x;
            this.y= y;
            this.z=z;
        }
        @Override
        public int compareTo(Tuple o){
            return this.x-o.x;
        }
    }

    static class Pair implements Comparable<Pair>{
        public int x;
        public int y;
        public Pair(int x, int y){
            this.x= x;
            this.y= y;
        }

        @Override
        public int compareTo(Pair o) {
            return this.x-o.x;
        }
    }

    // public static class compareL implements Comparator<Tuple>{
    //     @Override
    //     public int compare(Tuple t1, Tuple t2) {
    //         return t2.l - t1.l;
    //     }
    // }

    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public double nextDouble(){
            return Double.parseDouble(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
        public int[] nextIntArr(int n){
            int[] arr= new int[n];
            for(int i=0;i<n;i++) arr[i]= nextInt();
            return arr;
        }
        public long[] nextLongArr(int n){
            long[] arr= new long[n];
            for(int i=0;i<n;i++) arr[i]= nextLong();
            return arr;
        }
        public List<Integer> nextIntList(int n){
            List<Integer> arr= new ArrayList<>();
            for(int i=0;i<n;i++) arr.add(nextInt());
            return arr;
        }
        public int[][] nextIntMatArr(int n, int m){
            int[][] mat= new int[n][m];
            for(int i=0;i<n;i++) for(int j=0;j<m;j++) mat[i][j]= nextInt();
            return mat;
        }
        public List<List<Integer>> nextIntMatList(int n, int m){
            List<List<Integer>> mat= new ArrayList<>();
            for(int i=0;i<n;i++){
                List<Integer> temp= new ArrayList<>();
                for(int j=0;j<m;j++) temp.add(nextInt());
                mat.add(temp);
            }
            return mat;
        }
        public char[] nextStringCharArr(){
            return nextToken().toCharArray();
        }
    }
}