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;
const double pi = acos(-1.0);
const double eps = 1e-11;
void fastIO() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool isPalind[5001][5001];
int ans[5001];
int n, l, r;
string s;
void fillPalindDP() {
  for (int i = 0; i < n; i++) {
    l = i;
    r = i;
    while (l >= 0 && r < n && s[l] == s[r]) {
      isPalind[l][r] = true;
      l--;
      r++;
    }
    if (i < n - 1 && s[i] == s[i + 1]) {
      l = i;
      r = i + 1;
      while (l >= 0 && r < n && s[l] == s[r]) {
        isPalind[l][r] = true;
        l--;
        r++;
      }
    }
  }
}
int main() {
  fastIO();
  memset(isPalind, 0, sizeof(isPalind));
  cin >> s;
  n = int((s).size());
  fillPalindDP();
  int cK = 0;
  int tot, nj;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      tot = j - i + 1;
      cK = 0;
      nj = -1;
      if (isPalind[i][j]) {
        cK++;
        ans[cK]++;
        while (true) {
          nj = i + (tot / 2) - 1;
          if (isPalind[i][nj]) {
            tot = nj - i + 1;
            cK++;
            ans[cK]++;
          } else
            break;
          if (i == nj) {
            break;
          }
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}