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;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int z = power(a, b / 2);
  if (b % 2 == 0) return z * z;
  return a * z * z;
}
bool arr[5005][5005];
int main() {
  string s;
  cin >> s;
  int n = s.length();
  for (int i = 0; i < n; i++) arr[i][i] = true;
  for (int l = 2; l <= n; l++) {
    for (int i = 0; i < n - l + 1; i++) {
      int j = i + l - 1;
      if (l == 2)
        arr[i][j] = (s[i] == s[j]);
      else
        arr[i][j] = ((s[i] == s[j]) && arr[i + 1][j - 1]);
    }
  }
  int ans[n + 5];
  for (int i = 0; i <= n; i++) ans[i] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      int x = 0;
      int l = i, h = j;
      while (i <= j) {
        if (arr[l][h]) {
          ans[x]++;
          if ((h - l) % 2 == 0) {
            h = l + (h - l) / 2;
            h--;
          } else
            h = l + (h - l) / 2;
          x++;
        } else
          break;
      }
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  printf("\n");
}