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 main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  string s;
  cin >> s;
  int n = (int)s.size();
  for (int i = 0; i < n; ++i) {
    dp[i][i] = 1;
  }
  for (int len = 2; len <= n; ++len) {
    for (int l = 0; l + len <= n; ++l) {
      int r = l + len - 1;
      if (s[l] == s[r] && (l + 1 == r || dp[l + 1][r - 1]))
        dp[l][r] = dp[l][l + len / 2 - 1] + 1;
    }
  }
  vector<int> ans(n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (dp[i][j]) ans[dp[i][j] - 1]++;
  for (int i = n - 2; i >= 0; --i) ans[i] += ans[i + 1];
  for (auto it : ans) cout << it << ' ';
  cout << '\n';
}