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>
const int N = 5005;
int n, dp[N], ans[N];
char s[N];
struct pam {
  int tot, lst, ch[N][26], fail[N], len[N], cnt[N];
  int l, s[N];
  void init() {
    fail[1] = 0;
    fail[2] = 1;
    len[1] = -1;
    len[2] = 0;
    l = 0;
    s[0] = -1;
    tot = 2;
    lst = 1;
  }
  void ins(int c) {
    s[++l] = c;
    int p = lst;
    for (; s[l - len[p] - 1] != s[l]; p = fail[p])
      ;
    if (!ch[p][c]) {
      int q = ++tot;
      len[q] = len[p] + 2;
      if (p == 1)
        fail[q] = 2;
      else {
        int k = fail[p];
        for (; s[l - len[k] - 1] != s[l]; k = fail[k])
          ;
        fail[q] = ch[k][c];
      }
      ch[p][c] = q;
    }
    lst = ch[p][c];
    cnt[lst]++;
  }
  void build() {
    for (int j = tot; j > 2; --j) cnt[fail[j]] += cnt[j];
  }
} t;
int main() {
  scanf("%s", s);
  n = strlen(s);
  t.init();
  for (int j = 0; j < n; ++j) t.ins(s[j] - 'a');
  t.build();
  for (int i = 3; i <= t.tot; ++i) {
    dp[i] = 1;
    for (int j = t.fail[i]; j; j = t.fail[j])
      if (t.len[j] == t.len[i] / 2) {
        dp[i] = dp[j] + 1;
        break;
      }
    ans[dp[i]] += t.cnt[i];
  }
  for (int i = n; i; --i) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}