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 str[5212];
int f[5212][5212], ans[5212], len;
void Init(void) {
  scanf("%s", str + 1);
  len = strlen(str + 1);
  ans[1] = len;
  for (int i = 1; i <= len - 1; ++i) {
    f[i][i] = 1;
    if (str[i] == str[i + 1]) {
      f[i][i + 1] = 2;
      ++ans[2];
    }
  }
  f[len][len] = 1;
  return;
}
void Solve(void) {
  for (int i = 3; i <= len; ++i) {
    for (int j = 1; j + i - 1 <= len; ++j) {
      int k = j + i - 1;
      if (f[j + 1][k - 1] == false || str[j] != str[k]) continue;
      f[j][k] = f[j][j + i / 2 - 1] + 1;
      ++ans[f[j][k]];
    }
  }
  for (int i = len; i > 1; i--) ans[i - 1] += ans[i];
  for (int i = 1; i <= len; ++i) printf("%d ", ans[i]);
  return;
}
int main(void) {
  Init();
  Solve();
  return 0;
}