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, i, j, k, l;
string s;
bool res[10][5005][5005] = {};
long long int sum[5005] = {};
int po[20];
int main() {
  po[1] = 2;
  for (i = 2; i < 20; i++) po[i] = po[i - 1] * 2;
  cin >> s;
  n = s.length();
  for (i = 0; i < n; i++) {
    for (j = 1; i - j >= 0 && i + j < n; j++)
      if (s[j + i] != s[i - j])
        break;
      else {
        res[1][i - j][i + j] = 1;
        sum[1]++;
      }
    for (j = 0; i + j < n && i - j - 1 >= 0; j++)
      if (s[j + i] != s[i - j - 1])
        break;
      else {
        res[1][i - j - 1][i + j] = 1;
        sum[1]++;
      }
  }
  for (i = 0; i < n; i++) res[1][i][i] = 1;
  sum[1] += n;
  k = 2;
  bool rr = 0;
  int w;
  for (w = 2; w < 14; w++) {
    if (k == 9) {
      k = 4;
      rr = 1;
    }
    for (i = 0; i < n; i++)
      for (j = i - 1 + po[w - 1]; j < n; j++) {
        l = (j - i + 1) / 2;
        if (!rr) {
          if (res[k - 1][i][i + l - 1] && res[k - 1][j - l + 1][j] &&
              res[1][i][j]) {
            res[k][i][j] = 1;
            sum[w]++;
          } else
            res[k][i][j] = 0;
        } else {
          if (res[8][i][i + l - 1] && res[8][j - l + 1][j] && res[1][i][j]) {
            res[k][i][j] = 1;
            sum[w]++;
          } else
            res[k][i][j] = 0;
        }
      }
    if (rr) rr = 0;
    k++;
  }
  for (i = 1; i <= n; i++) cout << sum[i] << endl;
  return 0;
}