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 <iostream>
#include <vector>
#include <string>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); ++i)

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  string s; cin >> s;
  int N = s.size();
  vector<int> ans(N, 1);

  rep(i, N-1) {
    if(s[i] == 'R' && s[i+1] == 'R') {
      ans[i+2] += ans[i];
      ans[i] = 0;
    }
    if(s[N-1-i] == 'L' && s[N-1-i-1] == 'L') {
      ans[N-1-i-2] += ans[N-1-i];
      ans[N-1-i] = 0;
    }
  }

  rep(i, N) {
    cout << ans[i] << (i == N-1 ? '\n' : ' ');
  }
  return 0;
}
