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 <bits/stdc++.h>
using namespace std;

using ll=long long;
const ll MOD=1000000007;

int main() {
  string s;
  cin >> s;
  int k=s.size();
  vector<int> a(k);
  s +='R';
  a[k-1]==0;
  int n=-1,m,t;
  for(int i=0;i<k;i++){
    if(s[i]=='R'&&s[i+1]=='L'){
      t=i;
    }
    else
      a[i]==0;
    if(s[i]=='L'&&s[i+1]=='R'){
      m=i;
      a[t]=1+(m-t)/2+(t-n-1)/2;
      a[t+1]=m-n-a[t];
      n=m;
    }
  }
  for(int i=0;i<k-1;i++){
    cout << a[i] << " ";
  }
  cout << a[k-1] << endl;
}