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 = 32000;
const int mod = 1000000007;
const int inf = 2 * mod;
int dp[5001][5001];
pair<long long, long long> hashVal[5001];
pair<long long, long long> powVal[5001];
const long long base1 = 37;
const long long base2 = 2017;
const long long mod1 = 1e9 + 7;
const long long mod2 = 1e9 + 297;
int ans[50001];
string s;
inline int ceil(int p, int q) { return (p + q - 1) / q; }
int call(int l, int r) {
  if (l > r) return 0;
  if (l == r) return 1;
  if (l == r - 1) return (s[l] == s[r] ? 2 : 0);
  if (dp[l][r] != -1) return dp[l][r];
  int ret;
  if (s[l] != s[r] || call(l + 1, r - 1) == 0)
    ret = 0;
  else
    ret = 1 + call(l, ceil(l + r, 2) - 1);
  return dp[l][r] = ret;
}
int main() {
  cin >> s;
  int len = s.length();
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < len; i++) {
    for (int j = 0; j < len; j++) {
      ans[call(i, j)]++;
    }
  }
  for (int i = len - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= len; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}