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.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Calendar;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.SortedSet;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

/**
 * #
 *
 * @author pttrung
 */
public class C_Round_724_Div2 {
    public static long MOD = 1000000007;
    static long[][] dp;

    public static void main(String[] args) throws FileNotFoundException {
        // PrintWriter out = new PrintWriter(new FileOutputStream(new File(
        // "output.txt")));
        PrintWriter out = new PrintWriter(System.out);
        Scanner in = new Scanner();
        int T = in.nextInt();
        for (int z = 0; z < T; z++) {

            int n = in.nextInt();
            String data = in.next();
            int[] dp = new int[n];
            int[] d = new int[n];
            int[] k = new int[n];
            for (int i = 0; i < n; i++) {
                if (data.charAt(i) == 'D') {
                    d[i]++;
                } else {
                    k[i]++;
                }
                if (i > 0) {
                    d[i] += d[i - 1];
                    k[i] += k[i - 1];
                }
                if (d[i] == i + 1 || k[i] == i + 1) {
                    dp[i] = i + 1;
                } else {
                    int x = d[i];
                    int y = k[i];
                    int g = (int) gcd(x, y);
                    x /= g;
                    y /= g;
                    int re = 1;
                    for(int j = (x + y); j < (i + 1); j += (x + y)){
                        int a = d[i] - d[i - j];
                        int b = k[i] - k[i - j];
                        int v = (int)gcd(a, b);
                        if(a / v == x && b / v == y){
                            re = dp[i - j] + 1;
                            break;
                        }
                    }
                    dp[i] = re;
                }
            }
            for(int i : dp){
                out.print(i + " ");
            }
            out.println();
        }

        out.close();
    }


    static long abs(long a) {
        return a < 0 ? -a : a;
    }


    public static int[] KMP(String val) {
        int i = 0;
        int j = -1;
        int[] result = new int[val.length() + 1];
        result[0] = -1;
        while (i < val.length()) {
            while (j >= 0 && val.charAt(j) != val.charAt(i)) {
                j = result[j];
            }
            j++;
            i++;
            result[i] = j;
        }
        return result;

    }

    public static boolean nextPer(int[] data) {
        int i = data.length - 1;
        while (i > 0 && data[i] < data[i - 1]) {
            i--;
        }
        if (i == 0) {
            return false;
        }
        int j = data.length - 1;
        while (data[j] < data[i - 1]) {
            j--;
        }
        int temp = data[i - 1];
        data[i - 1] = data[j];
        data[j] = temp;
        Arrays.sort(data, i, data.length);
        return true;
    }

    public static int digit(long n) {
        int result = 0;
        while (n > 0) {
            n /= 10;
            result++;
        }
        return result;
    }

    public static double dist(long a, long b, long x, long y) {
        double val = (b - a) * (b - a) + (x - y) * (x - y);
        val = Math.sqrt(val);
        double other = x * x + a * a;
        other = Math.sqrt(other);
        return val + other;

    }

    public static class Point implements Comparable<Point> {

        int x;
        long y;

        public Point(int start, long end) {
            this.x = start;
            this.y = end;
        }


        @Override
        public int compareTo(Point o) {
            return Integer.compare(x, o.x);
        }
    }

    public static class FT {

        long[] data;

        FT(int n) {
            data = new long[n];
        }

        public void update(int index, long value) {
            while (index < data.length) {
                data[index] += value;
                index += (index & (-index));
            }
        }

        public long get(int index) {
            long result = 0;
            while (index > 0) {
                result += data[index];
                index -= (index & (-index));
            }
            return result;

        }
    }

    public static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    public static long pow(long a, int b) {
        if (b == 0) {
            return 1;
        }
        if (b == 1) {
            return a;
        }

        long val = pow(a, b / 2);
        if (b % 2 == 0) {
            return val * val;
        } else {
            return val * (val * a);

        }
    }

    static class Scanner {

        BufferedReader br;
        StringTokenizer st;

        public Scanner() throws FileNotFoundException {
            // System.setOut(new PrintStream(new BufferedOutputStream(System.out), true));
            br = new BufferedReader(new InputStreamReader(System.in));
            //br = new BufferedReader(new InputStreamReader(new FileInputStream(new File("input.txt"))));
        }

        public String next() {

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

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public String nextLine() {
            st = null;
            try {
                return br.readLine();
            } catch (Exception e) {
                throw new RuntimeException();
            }
        }

        public boolean endLine() {
            try {
                String next = br.readLine();
                while (next != null && next.trim().isEmpty()) {
                    next = br.readLine();
                }
                if (next == null) {
                    return true;
                }
                st = new StringTokenizer(next);
                return st.hasMoreTokens();
            } catch (Exception e) {
                throw new RuntimeException();
            }
        }
    }
}