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;
bool dp[5004][5004] = {0};
int val[5004][5004] = {0};
int ans[5005];
int main() {
  string str;
  cin >> str;
  str = "!" + str;
  for (int i = 1; i <= (int)str.length() - 1; i++)
    dp[i][i] = 1, val[i][i] = 1, ans[1]++;
  for (int l = 2; l <= (int)str.length() - 1; l++)
    for (int i = 1; i <= (int)str.length() - 1; i++) {
      int j = i + l - 1;
      if (j > str.length() - 1) break;
      if (l == 2) {
        if (str[i] == str[j]) dp[i][j] = 1;
        if (dp[i][j]) val[i][j] = val[i][i + l / 2 - 1] + 1, ans[val[i][j]]++;
        continue;
      }
      if (str[i] == str[j] && dp[i + 1][j - 1] == 1)
        dp[i][j] = 1;
      else
        dp[i][j] = 0;
      if (dp[i][j]) val[i][j] = val[i][i + l / 2 - 1] + 1, ans[val[i][j]]++;
    }
  for (int i = str.length() - 1; i >= (int)1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= (int)str.length() - 1; i++) cout << ans[i] << " ";
  return 0;
}