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;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, TT<T...>& c) {
  out << "{ ";
  for (auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const long long N = 5005, mod = 998244353;
string s;
long long dp[N][N], ans[N], n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s;
  n = s.size();
  for (long long len = 1; len <= n; len++) {
    for (long long l = 0; l + len - 1 < n; l++) {
      long long r = l + len;
      if (len == 1) {
        dp[l][r] = 1;
        continue;
      } else if (len == 2) {
        dp[l][r] = (s[l] == s[r - 1] ? 2 : 0);
        continue;
      }
      if (s[l] != s[r - 1] || !dp[l + 1][r - 1]) continue;
      dp[l][r] = 1;
      long long mid = (l + r) >> 1;
      if (len & 1) {
        if (dp[l][mid] && dp[mid + 1][r]) dp[l][r] = dp[l][mid] + 1;
      } else {
        if (dp[l][mid] && dp[mid][r]) dp[l][r] = dp[l][mid] + 1;
      }
    }
  }
  for (long long len = 1; len <= n; len++)
    for (long long l = 0; l + len - 1 < n; l++) ans[dp[l][l + len]]++;
  for (long long i = n - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
}