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>
int f[5000][5000], l;
int ans[5001];
char s[5005];
int main() {
  scanf("%s", s), l = strlen(s);
  for (int i = 0; i < l; i++)
    f[i][i] = 1, f[i][i + 1] = s[i] == s[i + 1] ? 2 : 0;
  for (int d = 2; d < l; d++)
    for (int i = 0, j = d; j < l; i++, j++) {
      if (s[i] != s[j] || f[i + 1][j - 1] == 0)
        f[i][j] = 0;
      else
        f[i][j] = f[i][i - 1 + (j - i + 1) / 2] + 1;
    }
  for (int i = 0; i < l; i++)
    for (int j = i; j < l; j++) ans[f[i][j]]++;
  for (int i = l - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= l; i++) printf("%d ", ans[i]);
  return 0;
}