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
#include <cstdio>
#include <cstring>
const int MAXN = 1e5+5;
int A[MAXN];
char S[MAXN];
int N;

int main() {
  scanf("%s", S);
  N = strlen(S);
  for (int i = 0; i < N; ++i) {
    A[i] = 0;
  }
  for (int i = 0, j; i < N; i = j) {
    int p = i;
    while (S[i] == 'R') ++i;
    j = i;
    while ((j < N) && (S[j] == 'L')) ++j;
    int lhs = i-p;
    int rhs = j-i;
    A[i] = lhs / 2 + (rhs + 1) / 2;
    A[i-1] = (lhs + 1) / 2 + rhs / 2;
  }
  for (int i = 0; i < N; ++i) {
    printf("%d%c", A[i], (i+1 == N) ? '\n' : ' ');
  }
  return 0;
}