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];
int ans[5005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  s = '$' + s;
  for (int i = 1; i <= n; ++i) dp[i][i] = 1;
  for (int i = 1; i < n; ++i) dp[i][i + 1] = s[i] == s[i + 1] ? 2 : 0;
  for (int l = 3; l <= n; ++l)
    for (int i = 1, j = i + l - 1; j <= n; i++, j++)
      dp[i][j] =
          s[i] != s[j] or !dp[i + 1][j - 1] ? 0 : dp[i][i + (l - 2) / 2] + 1;
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j) ans[dp[i][j]]++;
  for (int i = n; i >= 1; --i) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
  return 0;
}