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;
bool is_palin[5000][5000];
int palin_d[5000];
int result[5001];
int main(int argc, char** argv) {
  std::ios::sync_with_stdio(false);
  string s;
  cin >> s;
  int n = s.length();
  for (int i = 0; i <= n; ++i) {
    result[i] = 0;
  }
  for (int i = n - 1; i >= 0; --i) {
    for (int j = i; j < n; ++j) {
      is_palin[i][j] =
          ((i + 1 > j - 1) ? true : is_palin[i + 1][j - 1]) && (s[i] == s[j]);
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i; j < n; ++j) {
      if (i == j) {
        palin_d[j] = 1;
      } else {
        if (is_palin[i][j]) {
          palin_d[j] = palin_d[i + (j - i + 1) / 2 - 1] + 1;
        } else {
          palin_d[j] = 0;
        }
      }
      ++result[palin_d[j]];
    }
  }
  for (int i = n - 2; i >= 1; --i) {
    result[i] += result[i + 1];
  }
  for (int i = 1; i <= n; ++i) {
    cout << result[i];
    if (i < n) {
      cout << " ";
    }
  }
  cout << endl;
}