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;
const int MAX = 5e3 + 11;
int sz;
char str[MAX];
int pal[MAX][MAX];
int mem[MAX][MAX];
int ans[MAX];
int main() {
  scanf("%s", str + 1);
  int sz = strlen(str + 1);
  for (int i = 1; i <= sz; ++i)
    for (int l = 1; l + i - 1 <= sz; ++l) {
      int r = l + i - 1;
      pal[l][r] = (str[l] == str[r]) && (r - l <= 1 || (pal[l + 1][r - 1]));
      mem[l][r] = pal[l][r] ? mem[l][l + i / 2 - 1] + 1 : 0;
      ans[mem[l][r]]++;
    }
  for (int i = sz; i; --i) ans[i - 1] += ans[i];
  for (int i = 1; i <= sz; ++i) printf("%d\n", ans[i]);
  return 0;
}