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;
long long x, y, p[5005], f[5005], ans[5005][5005];
string s;
void solve() {
  cin >> s;
  for (__typeof(((long long)s.size())) i = (0); i < ((long long)s.size());
       i++) {
    x = y = 0;
    for (__typeof(((long long)s.size())) j = (i); j < ((long long)s.size());
         j++) {
      x = (x * 31 % 1000000007 + s[j] - 'a' + 1) % 1000000007;
      y = (y + (s[j] - 'a' + 1) * p[j - i] % 1000000007) % 1000000007;
      ans[i][j] = (x == y);
    }
  }
  for (__typeof(((long long)s.size())) l = (0); l < ((long long)s.size());
       l++) {
    for (__typeof(((long long)s.size() - l)) i = (0);
         i < ((long long)s.size() - l); i++) {
      long long j = i + l;
      if (ans[i][j] && ans[i][(i + j - 1) / 2] ==
                           ans[(i + j - 1) / 2 + 1 + (j - i + 1) % 2][j])
        ans[i][j] = ans[i][(i + j - 1) / 2] + 1;
      f[ans[i][j]]++;
    }
  }
  for (__typeof(((long long)s.size())) i = ((long long)s.size()); i >= (0); i--)
    f[i] += f[i + 1];
  for (__typeof(((long long)s.size() + 1)) i = (1);
       i < ((long long)s.size() + 1); i++)
    cout << f[i] << " ";
}
void prep() {
  p[0] = 1;
  for (__typeof((5005)) i = (1); i < (5005); i++)
    p[i] = p[i - 1] * 31 % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  prep();
  cout << fixed << setprecision(12);
  while (t--) solve();
  return 0;
}