Given is a string S consisting of `L` and `R`.

Let N be the length of S. There are N squares arranged from left to right, and the i-th character of S from the left is written on the i-th square from the left.

The character written on the leftmost square is always `R`, and the character written on the rightmost square is always `L`.

Initially, one child is standing on each square.

Each child will perform the move below 10^{100} times:

* Move one square in the direction specified by the character written in the square on which the child is standing. `L` denotes left, and `R` denotes right.



Find the number of children standing on each square after the children performed the moves.

Constraints

* S is a string of length between 2 and 10^5 (inclusive).
* Each character of S is `L` or `R`.
* The first and last characters of S are `R` and `L`, respectively.

Input

Input is given from Standard Input in the following format:


S


Output

Print the number of children standing on each square after the children performed the moves, in order from left to right.

Examples

Input

RRLRL


Output

0 1 2 1 1


Input

RRLLLLRLRRLL


Output

0 3 3 0 0 0 1 1 0 2 2 0


Input

RRRLLRLLRRRLLLLL


Output

0 0 3 2 0 2 1 0 0 0 4 4 0 0 0 0
import java.util.*;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		int N = S.length();
		int R_index = 0;
		int L_index = 0;
		int now_R = 0;
		int even_count = 0;
		int odd_count = 0;
		int[] ans = new int[N];
		int len = 0;
		int R_first=0;

		for (int i=0;i<N;i++) {
			if (S.charAt(i)=='R') {
				if (now_R==0) {
					// if ((i-R_index)%2==0) {
					// 	even_count++;
					// } else if((i-R_index)%2==1) {
					// 	odd_count++;
					// }
					len++;
				} else {
					now_R=0;
					// ans[L_index]+=even_count;
					// ans[R_index]+=odd_count;
					// odd_count=0;
					// even_count=1;
					if ((R_index-R_first)%2==0) {
						ans[R_index]+=len/2;
						if (len%2==1) {
							ans[R_index]++;
						}
						ans[L_index]+=len/2;
					} else {
						ans[L_index]+=len/2;
						if (len%2==1) {
							ans[L_index]++;
						}
						ans[R_index]+=len/2;
					}
					R_first=i;
					len=1;
				}
			} else if (S.charAt(i)=='L') {
				if (now_R==1) {
					// if ((i-L_index)%2==0) {
					// 	even_count++;
					// } else if((i-L_index)%2==1) {
					// 	odd_count++;
					// }
					len++;
				} else {
					now_R=1;
					R_index = i-1;
					L_index = i;
					len++;
					// if ((i-R_index%2)==0) {
					// 	ans[L_index]+=even_count;
					// 	ans[R_index]+=odd_count;
					// } else {
					// 	ans[L_index]+=odd_count;
					// 	ans[R_index]+=even_count;
					// }
					// odd_count=0;
					// even_count=1;
				}
			}
			if (i==N-1) {
				// ans[L_index]+=odd_count;
				// ans[R_index]+=even_count;
					if ((R_index-R_first)%2==0) {
						ans[R_index]+=len/2;
						if (len%2==1) {
							ans[R_index]++;
						}
						ans[L_index]+=len/2;
					} else {
						ans[L_index]+=len/2;
						if (len%2==1) {
							ans[L_index]++;
						}
						ans[R_index]+=len/2;
					}
			}
		}
		for (int i=0;i<N;i++) {
			if (i==N-1) {
				System.out.print(ans[i]);
			} else {
				System.out.print(ans[i] + " ");
			}
		}
	}
}