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 dp[5005][5005];
long long int cnt[5005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  string s;
  cin >> s;
  int i, j, k, l, r, mid, n = s.length();
  for (i = 0; i < n; i++) cnt[dp[i][i] = 1]++, dp[i + 1][i] = 1;
  for (i = 2; i <= n; i++)
    for (l = 0; l <= n - i; l++)
      cnt[dp[l][l + i - 1] = (s[l] == s[l + i - 1] and dp[l + 1][l + i - 2])
                                 ? (1 + dp[l][l + i / 2 - 1])
                                 : 0]++;
  for (i = n - 1; i; i--) cnt[i] += cnt[i + 1];
  for (i = 1; i <= n; i++) cout << cnt[i] << " ";
}