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 int maxn = 5005;
short f[maxn][maxn];
bool mark[maxn][maxn];
int ans[maxn];
string s;
int main() {
  cin >> s;
  for (int len = 1; len <= s.length(); len++) {
    for (int i = 0; i + len <= s.length(); i++) {
      int j = i + len - 1;
      if ((len == 1 || len == 2 || mark[i + 1][j - 1]) && s[i] == s[j]) {
        mark[i][j] = true;
        if (len == 1) {
          f[i][j] = 1;
        } else {
          f[i][j] = f[(i + j) / 2 + 1][j] + 1;
        }
        ans[f[i][j]]++;
      }
    }
  }
  for (int i = s.length(); i >= 1; i--) {
    ans[i] += ans[i + 1];
  }
  for (int i = 1; i <= s.length(); i++) {
    cout << ans[i] << ' ';
  }
  cout << endl;
}