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 int M = 1000000 + 10;
const int N = 5000 + 10;
char s[N];
int is[N][N], ans[N];
int deal(int l, int r) {
  if (l == r) return 1;
  if (!is[l][r]) return 0;
  return deal(l, l + (r - l + 1) / 2 - 1) + 1;
}
signed main() {
  scanf("%s", s + 1);
  int len = strlen(s + 1);
  for (int i = 1; i <= len; ++i) {
    is[i][i] = 1;
    if (i + 1 <= len && s[i] == s[i + 1]) is[i][i + 1] = 1;
  }
  for (int k = 3; k <= len; ++k) {
    for (int l = 1; l <= len - k + 1; ++l) {
      int r = l + k - 1;
      if (is[l + 1][r - 1] && s[l] == s[r]) is[l][r] = 1;
    }
  }
  for (int i = 1; i <= len; ++i) {
    for (int j = i; j <= len; ++j) {
      ans[deal(i, j)]++;
    }
  }
  for (int i = len - 1; i >= 1; --i) ans[i] += ans[i + 1];
  for (int i = 1; i <= len; ++i) printf("%d%c", ans[i], i == len ? '\n' : ' ');
  return 0;
}