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;
signed main() {
  ios_base::sync_with_stdio(0);
  string second;
  cin >> second;
  int n = second.size();
  vector<vector<int>> dp(n, vector<int>(n, -1));
  vector<int> ans(n, 0);
  for (int i = 0; i < n; i++) {
    dp[i][i] = 1;
    for (int j = 1; i - j >= 0 && i + j <= n - 1; j++) {
      if (second[i - j] != second[i + j]) break;
      dp[i - j][i + j] = 1;
      if (dp[i - j][i - 1] != -1) dp[i - j][i + j] = dp[i - j][i - 1] + 1;
    }
    for (int j = 1; i - j >= 0 && i + j - 1 <= n - 1; j++) {
      if (second[i - j] != second[i + j - 1]) break;
      dp[i - j][i + j - 1] = 1;
      if (dp[i - j][i - 1] != -1) dp[i - j][i + j - 1] = dp[i - j][i - 1] + 1;
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      for (int p = 0; p < dp[i][j]; p++) ans[p]++;
    }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}