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;
using lli = long long int;
using pii = pair<int, int>;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vlli = vector<long long int>;
using vpii = vector<pair<int, int>>;
const int M = 5010;
const int q = 71;
int n, m;
string s;
int lHash[M][M], rHash[M][M], cnt[M];
short pal[M][M];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.length();
  s = '#' + s;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      lHash[i][j] = ((lli)lHash[i][j - 1] * q + (lli)s[j]) % 1000000007;
  for (int i = n; i >= 1; i--)
    for (int j = i; j <= n; j++)
      rHash[i][j] = ((lli)rHash[i + 1][j] * q + (lli)s[i]) % 1000000007;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      m = (j - i - 1) / 2;
      if (lHash[i][i + m] == rHash[j - m][j]) pal[i][j] = pal[i][i + m] + 1;
      cnt[pal[i][j]]++;
    }
  for (int i = n - 1; i >= 1; i--) cnt[i] += cnt[i + 1];
  for (int i = 1; i <= n; i++) cout << cnt[i] << " ";
}