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.*;
public class Main
{
	public void tq()throws Exception
	{
		st=new StringTokenizer(bq.readLine());
		int tq=i();
		sb=new StringBuilder(1000000);
		o:
		while(tq-->0)
		{
			int n=i();
			String s=s();
			HashMap<Integer,Integer> h=new HashMap<>();
			int a=0;
			int b=0;
			int i=-1;
			int f[]=new int[n];
			for(char c:s.toCharArray())
			{
				i++;
				if(c=='D')a++;
				else b++;
				if(a==0||b==0)
				{
					f[i]=i+1;
					continue;
				}
				int v=gcd(a,b);
				v=(a/v)*1000000+(b/v);
				Integer aa=h.get(v);
				if(aa==null)
				{
					h.put(v,1);
					f[i]=1;
				}
				else
				{
					h.put(v,aa+1);
					f[i]=aa+1;
				}
				//s(f);
			}
			s(f);
		}
		p(sb);
	}
    
    long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;
    long maxl=Long.MAX_VALUE,minl=Long.MIN_VALUE;
    BufferedReader bq=new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;StringBuilder sb;
    public static void main(String[] a)throws Exception{new Main().tq();}
    int[] so(int ar[]){Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
    Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
    long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
    Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
    char[] so(char ar[])
    {Character r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
    Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
    void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.append(s);}
    void s(char s){sb.append(s);}void s(double s){sb.append(s);}
    void ss(){sb.append(' ');}void sl(String s){sb.append(s);sb.append("\n");}
    void sl(int s){sb.append(s);sb.append("\n");}
    void sl(long s){sb.append(s);sb.append("\n");}void sl(char s){sb.append(s);sb.append("\n");}
    void sl(double s){sb.append(s);sb.append("\n");}void sl(){sb.append("\n");}
    int l(int v){return 31-Integer.numberOfLeadingZeros(v);}
    long l(long v){return 63-Long.numberOfLeadingZeros(v);}
    int min(int a,int b){return a<b?a:b;}
    int min(int a,int b,int c){return a<b?a<c?a:c:b<c?b:c;}
    int max(int a,int b){return a>b?a:b;}
    int max(int a,int b,int c){return a>b?a>c?a:c:b>c?b:c;}
    long min(long a,long b){return a<b?a:b;}
    long min(long a,long b,long c){return a<b?a<c?a:c:b<c?b:c;}
    long max(long a,long b){return a>b?a:b;}
    long max(long a,long b,long c){return a>b?a>c?a:c:b>c?b:c;}
    int abs(int a){return Math.abs(a);}
    long abs(long a){return Math.abs(a);}
    int sq(int a){return (int)Math.sqrt(a);}long sq(long a){return (long)Math.sqrt(a);}
    long gcd(long a,long b){while(b>0l){long c=a%b;a=b;b=c;}return a;}
    int gcd(int a,int b){while(b>0){int c=a%b;a=b;b=c;}return a;}
    boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=s.charAt(j--))return false;return true;}
    boolean[] si(int n)
    {boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]=true;for(int x=4;x<=n;x+=2)bo[x]=true;
    for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;y+=vv)bo[y]=true;}}
    return bo;}long mul(long a,long b,long m) 
    {long r=1l;a%=m;while(b>0){if((b&1)==1)r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}
    int i()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    return Integer.parseInt(st.nextToken());}
    long l()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    return Long.parseLong(st.nextToken());}String s()throws IOException
    {if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return st.nextToken();}
    double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    return Double.parseDouble(st.nextToken());}void p(Object p){System.out.print(p);}
    void p(String p){System.out.print(p);}void p(int p){System.out.print(p);}
    void p(double p){System.out.print(p);}void p(long p){System.out.print(p);}
    void p(char p){System.out.print(p);}void p(boolean p){System.out.print(p);}
    void pl(Object p){System.out.println(p);}void pl(String p){System.out.println(p);}
    void pl(int p){System.out.println(p);}void pl(char p){System.out.println(p);}
    void pl(double p){System.out.println(p);}void pl(long p){System.out.println(p);}
    void pl(boolean p){System.out.println(p);}void pl(){System.out.println();}
    void s(int a[]){for(int e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
    void s(long a[]){for(long e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
    void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}
    void s(char a[]){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
    void s(char ar[][]){for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}
    int[] ari(int n)throws IOException
    {int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    for(int x=0;x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}
    int[][] ari(int n,int m)throws IOException
    {int ar[][]=new int[n][m];for(int x=0;x<n;x++){if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}
    long[] arl(int n)throws IOException
    {long ar[]=new long[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    for(int x=0;x<n;x++) ar[x]=Long.parseLong(st.nextToken());return ar;}
    long[][] arl(int n,int m)throws IOException
    {long ar[][]=new long[n][m];for(int x=0;x<n;x++)
    {if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
    String[] ars(int n)throws IOException
    {String ar[]=new String[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    for(int x=0;x<n;x++) ar[x]=st.nextToken();return ar;}
    double[] ard(int n)throws IOException
    {double ar[]=new double[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}
    double[][] ard(int n,int m)throws IOException
    {double ar[][]=new double[n][m];for(int x=0;x<n;x++)
    {if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    for(int y=0;y<m;y++)ar[x][y]=Double.parseDouble(st.nextToken());}return ar;}
    char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());
    for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}
    char[][] arc(int n,int m)throws IOException{char ar[][]=new char[n][m];
    for(int x=0;x<n;x++){String s=bq.readLine();for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}
    void p(int ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
    for(int a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
    void p(int ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
    for(int a[]:ar){for(int aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
    void p(long ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
    for(long a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
    void p(long ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
    for(long a[]:ar){for(long aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
    void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;StringBuilder sb=new StringBuilder(c);
    for(String a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
    void p(double ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
    for(double a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
    void p(double ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
    for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
    void p(char ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
    for(char aa:ar){sb.append(aa);sb.append(' ');}System.out.println(sb);}
    void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
    for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
}