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 N = 5002;
int dp[N][N], pallindrome[N][N];
string s;
bool pal(int l, int r) {
  if (l > r) return 0;
  if (l == r) return 1;
  if (r - l == 1) return s[l] == s[r];
  if (pallindrome[l][r] != -1) return pallindrome[l][r];
  if (s[l] != s[r]) return pallindrome[l][r] = 0;
  return pallindrome[l][r] = pal(l + 1, r - 1);
}
int go(int l, int r) {
  if (l == r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  int k = (r - l) / 2 - (r - l + 1) % 2;
  int ans = 0;
  if (pal(l, r)) ans = 1;
  if (pal(l, r) && pal(l, l + k) && pal(r - k, r))
    ans = 1 + min(go(l, l + k), go(r - k, r));
  return dp[l][r] = ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  int n = s.length();
  vector<int> ans(n + 2, 0);
  memset(pallindrome, -1, sizeof pallindrome);
  memset(dp, -1, sizeof dp);
  for (int i = 0; i < n; ++i)
    for (int j = i; j < n; ++j) {
      ans[go(i, j)]++;
    }
  for (int i = n; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  return 0;
}