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 mul = 131, p = 1000000007;
char s[5010];
int pref[5010], suff[5010], ans[5010][5010], mi[5010], cnt[5010], out[5010];
int pHash(int l, int r) {
  return (pref[r] - (long long)pref[l - 1] * mi[r - l + 1] % p + p) % p;
}
int sHash(int l, int r) {
  return (suff[l] - (long long)suff[r + 1] * mi[r - l + 1] % p + p) % p;
}
int main() {
  scanf("%s", s + 1);
  int l = strlen(s + 1);
  mi[0] = 1;
  for (int i = 1; i <= l; i++) mi[i] = (long long)mi[i - 1] * mul % p;
  for (int i = 1; i <= l; i++)
    pref[i] = ((long long)pref[i - 1] * mul + s[i] - 'a') % p;
  for (int i = l; i; i--)
    suff[i] = ((long long)suff[i + 1] * mul + s[i] - 'a') % p;
  for (int i = 1; i <= l; i++) ans[i][i] = 1;
  for (int i = 2; i <= l; i++)
    for (int j = 1; j + i - 1 <= l; j++)
      if (pHash(j, j + i - 1) == sHash(j, j + i - 1))
        ans[j][j + i - 1] = ans[j][j + i / 2 - 1] + 1;
  for (int i = 1; i <= l; i++)
    for (int j = i; j <= l; j++) cnt[ans[i][j]]++;
  for (int i = l; i; i--) out[i] = out[i + 1] + cnt[i];
  for (int i = 1; i <= l; i++) printf("%d%c", out[i], i == l ? '\n' : ' ');
  return 0;
}