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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(0);
  string s;
  cin >> s;
  int n = (s).size();
  vector<vector<int> > dp(n + 10, vector<int>(n + 10));
  for (int len = 1; len <= n; ++len) {
    for (int i = 0; i < n - len + 1; ++i) {
      int j = i + len - 1;
      int mid = (i + j + 1) / 2;
      if (j == i) {
        dp[i][j] = 1;
      } else if (i + 1 == j) {
        dp[i][j] = (s[i] == s[j]);
        dp[i][j] *= 2;
      } else {
        if (s[i] == s[j] && dp[i + 1][j - 1]) {
          dp[i][j] = dp[i][mid - 1] + 1;
        } else {
          dp[i][j] = 0;
        }
      }
    }
  }
  vector<int> v(n + 1, 0);
  for (int i = 0; i < n + 1; ++i) {
    for (int j = 0; j < n + 1; ++j) {
      ++v[dp[i][j]];
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    v[i] += v[i + 1];
  }
  for (int i = 1; i <= n; ++i) {
    cout << v[i] << " ";
  }
  return 0;
}