Problem: Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes.

A string is 1-palindrome if and only if it reads the same backward as forward.

A string is k-palindrome (k > 1) if and only if: 

  1. Its left half equals to its right half. 
  2. Its left and right halfs are non-empty (k - 1)-palindromes. 



The left half of string t is its prefix of length ⌊|t| / 2⌋, and right half — the suffix of the same length. ⌊|t| / 2⌋ denotes the length of string t divided by 2, rounded down.

Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times.

Input

The first line contains the string s (1 ≤ |s| ≤ 5000) consisting of lowercase English letters.

Output

Print |s| integers — palindromic characteristics of string s.

Examples

Input

abba


Output

6 1 0 0 


Input

abacaba


Output

12 4 1 0 0 0 0 

Note

In the first example 1-palindromes are substring «a», «b», «b», «a», «bb», «abba», the substring «bb» is 2-palindrome. There are no 3- and 4-palindromes here.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int arr[5555][5555], pp[5555];
string a;
int main(void) {
  cin >> a;
  for (int e = 0; e < a.size(); e++) {
    for (int p = e; p >= 0; p--) {
      if (a[e] == a[p]) {
        if (e - 1 < p + 1) {
          arr[p][e] = 1;
          if (p + 1 == e) {
            arr[p][e] = 2;
          }
        } else {
          if (arr[p + 1][e - 1]) {
            arr[p][e] = 1;
            int kk = (e - p + 1) / 2;
            if (arr[p][p + kk - 1] == arr[e - kk + 1][e]) {
              arr[p][e] = arr[p][p + kk - 1] + 1;
            }
          }
        }
      }
    }
  }
  int n = 0;
  for (int e = 0; e < a.size(); e++) {
    for (int p = 0; p < a.size(); p++) {
      for (int q = 1; q <= arr[e][p]; q++) pp[q] += 1;
    }
  }
  for (int e = 1; e <= a.size(); e++) cout << pp[e] << " ";
}