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 = 5010;
int f[maxn][maxn], n, ans[maxn];
char s[maxn];
bool h[maxn][maxn];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) {
    f[i][i] = h[i][i] = h[i][i - 1] = 1;
    ans[1]++;
  }
  for (int p = 2; p <= n; p++) {
    for (int i = 1; i + p - 1 <= n; i++) {
      int j = i + p - 1;
      h[i][j] = h[i + 1][j - 1] & (s[i] == s[j]);
      f[i][j] = h[i][j] ? f[i][(i + j - 1) / 2] + 1 : 0;
      ans[f[i][j]]++;
    }
  }
  for (int i = n; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}