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 p = 31;
const int mod = 1e9 + 9;
const int maxn = 5e3 + 10;
long long p_pow[maxn], h[2][maxn];
int n;
void init() {
  memset(p_pow, 0, sizeof(p_pow));
  p_pow[0] = 1;
  memset(h, 0, sizeof(h));
  for (int i = 1; i < maxn; i++) p_pow[i] = (p_pow[i - 1] * p) % mod;
}
void computeh(string &s, int tipo) {
  for (int i = 0; i < n; i++) {
    h[tipo][i + 1] = (h[tipo][i] + (s[i] - 'a' + 1) * p_pow[i]) % mod;
  }
}
long long sub_h(int lo, int hi, int tipo) {
  long long ans = (h[tipo][hi + 1] + mod - h[tipo][lo]) % mod;
  ans = (ans * p_pow[n - lo - 1]) % mod;
  return ans;
}
long long dp[maxn][maxn];
int main() {
  init();
  string s;
  cin >> s;
  n = s.size();
  computeh(s, 0);
  reverse(s.begin(), s.end());
  computeh(s, 1);
  vector<long long> ans(n + 1, 0);
  memset(dp, 0, sizeof(dp));
  ans[1] = n;
  for (int i = 0; i <= n; ++i) dp[i][i] = 1;
  for (int i = 0; i < s.size(); ++i) {
    for (int len = 2; i + len - 1 < n; ++len) {
      int hi = i + len - 1;
      long long sub1 = sub_h(i, hi, 0);
      long long sub2 = sub_h(n - 1 - hi, n - i - 1, 1);
      if (sub1 == sub2) {
        dp[i][hi] = dp[i][(i + hi - 1) / 2] + 1;
        for (long long k = dp[i][hi]; k >= 0; --k) ans[k]++;
      }
    }
  }
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  return 0;
}