There are n + 1 cities, numbered from 0 to n. n roads connect these cities, the i-th road connects cities i - 1 and i (i ∈ [1, n]).

Each road has a direction. The directions are given by a string of n characters such that each character is either L or R. If the i-th character is L, it means that the i-th road initially goes from the city i to the city i - 1; otherwise it goes from the city i - 1 to the city i.

A traveler would like to visit as many cities of this country as possible. Initially, they will choose some city to start their journey from. Each day, the traveler must go from the city where they currently are to a neighboring city using one of the roads, and they can go along a road only if it is directed in the same direction they are going; i. e., if a road is directed from city i to the city i + 1, it is possible to travel from i to i + 1, but not from i + 1 to i. After the traveler moves to a neighboring city, all roads change their directions to the opposite ones. If the traveler cannot go from their current city to a neighboring city, their journey ends; it is also possible to end the journey whenever the traveler wants to.

The goal of the traveler is to visit as many different cities as possible (they can visit a city multiple times, but only the first visit is counted). For each city i, calculate the maximum number of different cities the traveler can visit during exactly one journey if they start in the city i. 

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Each test case consists of two lines. The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5). The second line contains the string s consisting of exactly n characters, each character is either L or R.

It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.

Output

For each test case, print n + 1 integers. The i-th integer should be equal to the maximum number of different cities the traveler can visit during one journey if this journey starts in the i-th city.

Example

Input


2
6
LRRRLL
3
LRL


Output


1 3 2 3 1 3 2
1 4 1 4
//Utilities
import java.io.*;
import java.util.*;

public class Main {
	static int T;
	static int N;
	static char[] dir;
	static int[] res;
	static int r, cnt;
	
	public static void main(String[] args) throws IOException {
		T = in.iscan();
		while (T-- > 0) {
			N = in.iscan();
			dir = in.sscan().toCharArray();
			res = new int[N+1];
			// even indexes
			for (int i = 0; i <= N; i++) {
				if (i % 2 == 1) continue;
				r = i;
				dfs(i, i);
				cnt = r - i + 1;
				if (i-1 >= 0 && dir[i-1] == 'L') cnt++;
				for (int j = i; j <= r; j++) {
					if (j % 2 == 0) res[j] = cnt;
				}
				i = r;
			}
			// odd indexes
			for (int i = 0; i <= N; i++) {
				if (i % 2 == 0) continue;
				r = i;
				dfs(i, i);
				cnt = r - i + 1;
				if (i-1 >= 0 && dir[i-1] == 'L') cnt++;
				for (int j = i; j <= r; j++) {
					if (j % 2 == 1) res[j] = cnt;
				}
				i = r;
			}
			for (int i = 0; i <= N; i++) {
				out.print(res[i] + " ");
			}
			out.println();
		}
		out.close();
	} 
	
	static void dfs(int st, int idx) {
		if (idx == N) return;
		int d = dir[idx] == 'L' ? 0 : 1;
		if ((idx - st) % 2 == 1) {
			d ^= 1;
		}
		if (d == 1) {
			r = idx+1;
			dfs(st, idx+1);
		}
	}
	
	static INPUT in = new INPUT(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	private static class INPUT {

		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar, numChars;

		public INPUT (InputStream stream) {
			this.stream = stream;
		}

		public INPUT (String file) throws IOException {
			this.stream = new FileInputStream (file);
		}

		public int cscan () throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read (buf);
			}
			
			if (numChars == -1)
				return numChars;

			return buf[curChar++];
		}

		public int iscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			int res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public String sscan () throws IOException {
			int c = cscan ();
			
			while (space (c))
				c = cscan ();

			StringBuilder res = new StringBuilder ();

			do {
				res.appendCodePoint (c);
				c = cscan ();
			}
			while (!space (c));

			return res.toString ();
		}

		public double dscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			double res = 0;

			while (!space (c) && c != '.') {
				if (c == 'e' || c == 'E')
					return res * UTILITIES.fast_pow (10, iscan ());
				
				res *= 10;
				res += c - '0';
				c = cscan ();
			}

			if (c == '.') {
				c = cscan ();
				double m = 1;

				while (!space (c)) {
					if (c == 'e' || c == 'E')
						return res * UTILITIES.fast_pow (10, iscan ());

					m /= 10;
					res += (c - '0') * m;
					c = cscan ();
				}
			}

			return res * sgn;
		}

		public long lscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			long res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public boolean space (int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

	public static class UTILITIES {

		static final double EPS = 10e-6;

		public static int lower_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] >= x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int upper_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] > x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

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

		public static int lcm (int a, int b) {
			return a * b / gcd (a, b);
		}

		public static long fast_pow_mod (long b, long x, int mod) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;

			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
		}

		public static int fast_pow (int b, int x) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow (b * b, x / 2);

			return b * fast_pow (b * b, x / 2);
		}

		public static long choose (long n, long k) {
			k = Math.min (k, n - k);
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = val * (n - i) / (i + 1);

			return val;
		}

		public static long permute (int n, int k) {
			if (n < k) return 0;
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = (val * (n - i));

			return val;
		}
	}
}
