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;
vector<vector<int> > dp;
vector<int> ans;
int main() {
  string s;
  cin >> s;
  dp.resize(s.size(), vector<int>(s.size(), 0));
  ans.resize(s.size() + 1, 0);
  for (int i = 0; i < s.size(); ++i) {
    dp[i][i] = 1;
    ++ans[dp[i][i]];
  }
  for (int k = 2; k < s.size() + 1; ++k)
    for (int l = 0, r = l + k - 1; r < s.size(); ++l, ++r) {
      if (s[l] == s[r] && (k < 3 || dp[l + 1][r - 1])) {
        dp[l][r] += dp[l][l - 1 + k / 2] + 1;
        ++ans[dp[l][r]];
      }
    }
  for (int i = s.size() - 1; i >= 1; --i) ans[i] += ans[i + 1];
  for (int i = 1; i < ans.size(); ++i) cout << ans[i] << ' ';
  return 0;
}