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.math.BigInteger;
import java.util.*;

public class Main {
    static int MOD = 1000000007;

    // After writing solution, quick scan for:
    //   array out of bounds
    //   special cases e.g. n=1?
    //   npe, particularly in maps
    //
    // Big numbers arithmetic bugs:
    //   int overflow
    //   sorting, or taking max, after MOD
    void solve() throws IOException {
        int T = ri();
        for (int Ti = 0; Ti < T; Ti++) {
            int n = ri();
            char[] s = rs();

            // D : K
            int[] dprefix = new int[n];
            int[] kprefix = new int[n];
            dprefix[0] = s[0] == 'D' ? 1 : 0;
            kprefix[0] = s[0] == 'K' ? 1 : 0;
            for (int i = 1; i < n; i++) {
                dprefix[i] = dprefix[i-1] + (s[i] == 'D' ? 1 : 0);
                kprefix[i] = kprefix[i-1] + (s[i] == 'K' ? 1 : 0);
            }

            Map<Pair, Integer> ratioToIdx = new HashMap<>();
            Pair init = new Pair(dprefix[0], kprefix[0]);
            ratioToIdx.put(init, 0);

            // dp[i] is the best answer for the prefix dp[0..i]
            int[] dp = new int[n];
            dp[0] = 1;
            pw.print(1 + " ");

            // maybe remember the ratio for each s[0..i].
            // when we encounter some ratio, we look up a map for the last time that ratio occurred.

            for (int i = 1; i < n; i++) {
                // Overall ratio is dprefix[i] : kprefix[i].
                // As long as the first/last subsegment has the same ratio, we're guaranteed the rest does as well
                // Thus, the O(n^2) solution is clear.
                //   - compute the ratio D : K for the string
                //   - greedily take characters while that ratio is maintained.
                //
                // We need to make it faster...
                //
                // If we take the first/last segment, dp could be used to remember the answer to the rest.
                // So now the matter is: how do we find the last segment that works?
                int D = dprefix[i];
                int K = kprefix[i];
                if (D == 0 || K == 0) {
                    dp[i] = i+1;
                    pw.print(i+1 + " ");
                    continue;
                }
                int g = gcd(D, K);
                Pair curr = new Pair(D / g, K / g);
                if (g == 1) {
                    dp[i] = 1;
                    pw.print("1 ");
                    ratioToIdx.put(curr, i);
                    continue;
                }
                
                if (!ratioToIdx.containsKey(curr)) {
                    dp[i] = 1;
                    pw.print("1 ");
                    ratioToIdx.put(curr, i);
                    continue;
                } else {
                    int lastSeen = ratioToIdx.get(curr);
                    dp[i] = dp[lastSeen] + 1;
                    pw.print(dp[i] + " ");
                    ratioToIdx.put(curr, i);
                    continue;
                }
            }
            pw.println();
        }
    }
    // IMPORTANT
    // DID YOU CHECK THE COMMON MISTAKES ABOVE?

    public static int gcd(int a, int b) {
        return a == 0 ? b : gcd(b % a, a);
    }

    // Template code below

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        m.solve();
        m.close();
    }

    void close() throws IOException {
        pw.flush();
        pw.close();
        br.close();
    }

    int ri() throws IOException {
        return Integer.parseInt(br.readLine().trim());
    }

    long rl() throws IOException {
        return Long.parseLong(br.readLine().trim());
    }

    int[] ril(int n) throws IOException {
        int[] nums = new int[n];
        int c = 0;
        for (int i = 0; i < n; i++) {
            int sign = 1;
            c = br.read();
            int x = 0;
            if (c == '-') {
                sign = -1;
                c = br.read();
            }
            while (c >= '0' && c <= '9') {
                x = x * 10 + c - '0';
                c = br.read();
            }
            nums[i] = x * sign;
        }
        while (c != '\n' && c != -1) c = br.read();
        return nums;
    }

    long[] rll(int n) throws IOException {
        long[] nums = new long[n];
        int c = 0;
        for (int i = 0; i < n; i++) {
            int sign = 1;
            c = br.read();
            long x = 0;
            if (c == '-') {
                sign = -1;
                c = br.read();
            }
            while (c >= '0' && c <= '9') {
                x = x * 10 + c - '0';
                c = br.read();
            }
            nums[i] = x * sign;
        }
        while (c != '\n' && c != -1) c = br.read();
        return nums;
    }

    int[] rkil() throws IOException {
        int sign = 1;
        int c = br.read();
        int x = 0;
        if (c == '-') {
            sign = -1;
            c = br.read();
        }
        while (c >= '0' && c <= '9') {
            x = x * 10 + c - '0';
            c = br.read();
        }
        return ril(x);
    }

    long[] rkll() throws IOException {
        int sign = 1;
        int c = br.read();
        int x = 0;
        if (c == '-') {
            sign = -1;
            c = br.read();
        }
        while (c >= '0' && c <= '9') {
            x = x * 10 + c - '0';
            c = br.read();
        }
        return rll(x);
    }

    char[] rs() throws IOException {
        return br.readLine().toCharArray();
    }

    void sort(int[] A) {
        Random r = new Random();
        for (int i = A.length-1; i > 0; i--) {
            int j = r.nextInt(i+1);
            int temp = A[i];
            A[i] = A[j];
            A[j] = temp;
        }
        Arrays.sort(A);
    }

    void printDouble(double d) {
        pw.printf("%.16f", d);
    }
}

class Pair {
    int fst;
    int snd;
    Pair(int f, int s) {
        fst = f;
        snd = s;
    }
    public int hashCode() {
        return fst * 9991 + snd;
    }
    public boolean equals(Object o) {
        if (!(o instanceof Pair)) return false;
        Pair p = (Pair) o;
        return fst == p.fst && snd == p.snd;
    }
}