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;
string s;
int n;
int DP[5001][5001];
int ans[5001];
void kpalindromo() {
  for (int len = 1; len < (int)n + 1; ++len) {
    for (int l = 0; l < (int)n - len + 1; ++l) {
      int r = l + len, m = (l + r) >> 1;
      if (len == 1)
        DP[l][r] = 1;
      else if (len == 2)
        DP[l][r] = (s[l] == s[r - 1] ? 2 : 0);
      else if (s[l] != s[r - 1] || DP[l + 1][r - 1] == 0) {
        DP[l][r] = 0;
      } else {
        DP[l][r] = DP[l][m] + 1;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = (int)s.size();
  kpalindromo();
  for (int len = 1; len <= n; len++) {
    for (int l = 0; l < n - len + 1; l++) {
      ans[DP[l][l + len]]++;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    ans[i] += ans[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
  return 0;
}