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;
int n, m;
int f[5010][5010], g[5010][5010], ans[5010];
string s;
int main() {
  cin >> s;
  int n = (int)s.length();
  for (int i = 0; i < n; i++) {
    f[i][i] = 1;
    g[i][i] = 1;
  }
  for (int i = n - 2; i >= 0; i--) {
    for (int j = i + 1; j < n; j++) {
      if (s[i] == s[j] && (i + 1 == j || g[i + 1][j - 1])) {
        g[i][j] = 1;
        f[i][j] = f[i][i + (j - i + 1) / 2 - 1] + 1;
      }
    }
  }
  memset(ans, 0, sizeof ans);
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (g[i][j]) ans[f[i][j]]++;
    }
  }
  for (int i = n - 2; i >= 0; i--) {
    ans[i] += ans[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}