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 dp[5050][5050];
int ans[5050];
int main() {
  ios::sync_with_stdio(0);
  cout.tie(NULL);
  cin.tie(NULL);
  string s;
  cin >> s;
  int i, j;
  for (i = 1; i <= s.length(); i++)
    for (j = 0; j < s.length() - i + 1; j++) {
      if (i == 1) {
        dp[j][j + i] = 1;
      } else if (i == 2) {
        dp[j][j + i] = ((s[j] == s[j + 1]) ? 2 : 0);
      } else {
        if (s[j] != s[j + i - 1] || !dp[j + 1][j + i - 1]) {
          continue;
        } else {
          int m = (j + j + i) / 2;
          dp[j][j + i] = 1;
          if (i % 2) {
            if (dp[j][m] && dp[m + 1][j + i]) {
              dp[j][j + i] = dp[j][m] + 1;
            }
          } else {
            if (dp[j][m] && dp[m][j + i]) {
              dp[j][j + i] = dp[j][m] + 1;
            }
          }
        }
      }
    }
  for (i = 1; i <= s.length(); i++)
    for (j = 0; j < s.length() - i + 1; j++) ans[dp[j][j + i]]++;
  for (i = s.length() - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (i = 1; i <= s.length(); i++) cout << ans[i] << ' ';
  cout << "\n";
}