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;
void solvre() {}
int main() {
  string s;
  cin >> s;
  int tam = s.length();
  int ans[5050] = {0};
  vector<int> pd[5050];
  for (int i = 0; i < tam; i++) {
    pd[i].resize(tam + 1 - i);
    pd[i][0] = 1;
    pd[i][1] = 1;
  }
  for (int j = 2; j <= tam; j++) {
    for (int i = 0; i <= tam - j; i++) {
      if (pd[i + 1][j - 2] > 0 && s[i] == s[i + j - 1]) {
        pd[i][j] = min(pd[i][j / 2], pd[i + (int)ceil(j / 2.0)][j / 2]) + 1;
      } else
        pd[i][j] = 0;
    }
  }
  for (int i = 0; i < tam; i++) {
    for (int j = 1; j <= tam - i; j++) {
      ans[pd[i][j]]++;
    }
  }
  for (int i = tam; i > 0; i--) {
    ans[i] += ans[i + 1];
  }
  for (int i = 1; i <= tam; i++) {
    if (i != 1) cout << " ";
    cout << ans[i];
  }
  cout << endl;
}