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;
const double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = s.length();
  int dp[n + 5][n + 5], cnt[n + 5];
  for (int i = 0; i < (n + 5); i++) {
    cnt[i] = 0;
    for (int j = 0; j < (n + 5); j++) {
      dp[i][j] = 0;
    }
  }
  for (int i = (n - 1); i >= 0; i--) {
    for (int j = (i); j < (n); j++) {
      if (i == j)
        dp[i][j] = 1;
      else if (i == j - 1) {
        if (s[i] == s[j]) dp[i][j] = 2;
      } else {
        if (!(s[i] != s[j] or dp[i + 1][j - 1] == 0))
          dp[i][j] = dp[i][(i + j - 1) / 2] + 1;
      }
    }
  }
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (n); j++) {
      cnt[dp[i][j]]++;
    }
  }
  for (int i = (n - 1); i >= 1; i--) cnt[i] += cnt[i + 1];
  for (int i = (1); i < (n + 1); i++) cout << cnt[i] << " ";
}