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;
char s[5010];
int a[5010][5010];
int num[5010];
int main() {
  int i, j, n, len, l, r, mid;
  scanf("%s", s);
  memset(a, 0, sizeof(a));
  len = strlen(s);
  for (i = 0; i < len; i++) {
    a[i][i] = 1;
    for (l = i - 1, r = i + 1; l >= 0 && r < len; l--, r++) {
      if (s[l] == s[r])
        a[l][r] = 1;
      else
        break;
    }
    for (l = i, r = i + 1; l >= 0 && r < len; l--, r++) {
      if (s[l] == s[r])
        a[l][r] = 1;
      else
        break;
    }
  }
  for (i = 1; i < len; i++) {
    for (l = 0; l + i < len; l++)
      if (a[l][l + i] == 1) {
        r = l + i;
        mid = l + i / 2;
        if (i % 2 == 0) {
          if (a[l][mid - 1] == a[mid + 1][r] && a[l][mid - 1] > 0)
            a[l][r] = a[l][mid - 1] + 1;
        } else {
          if (a[l][mid] == a[mid + 1][r] && a[l][mid] > 0)
            a[l][r] = a[l][mid] + 1;
        }
      }
  }
  memset(num, 0, sizeof(num));
  for (i = 0; i < len; i++)
    for (j = i; j < len; j++)
      if (a[i][j] > 0) num[a[i][j]]++;
  for (i = len - 1; i >= 1; i--) num[i] += num[i + 1];
  for (i = 1; i <= len; i++) printf("%d ", num[i]);
  printf("\n");
  return 0;
}