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 p[5000];
char c[5000][5000];
string s;
void pv(int i, int j, int v) {
  c[i][j] = v;
  p[v - 1]++;
}
void lr(int i, int j) {
  int b = i;
  while (i >= 0 && j < s.size() && s[i] == s[j]) {
    pv(i, j, c[i][b] + 1);
    i--;
    j++;
  }
}
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    pv(i, i, 1);
    lr(i - 1, i + 1);
    lr(i, i + 1);
  }
  for (int i = s.size() - 1; i > 0; i--) {
    p[i - 1] += p[i];
  }
  for (int i = 0; i < s.size(); i++) {
    cout << p[i] << " ";
  }
}