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;
string s;
long long dp[5001][5001];
long long ans[5001];
long long h[5001];
long long pw[5001];
long long p = 1e9 + 7, q = 321;
void calc() {
  pw[0] = 1;
  for (long long i = 1; i <= 5000; ++i) pw[i] = (pw[i - 1] * q) % p;
  for (long long i = 1; i <= (long long)((s).size()); ++i)
    h[i] = (h[i - 1] * q + (s[i - 1])) % p;
}
long long get(long long l, long long r) {
  return (h[r + 1] - h[l] * pw[r - l + 1] + p * p) % p;
}
long long ok(long long l, long long len) {
  return get(l, l + len / 2 - 1) == get(l + (len + 1) / 2, l + len - 1);
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> s;
  for (long long i = 0; i < (long long)((s).size()); ++i) ans[i] = 0;
  calc();
  for (long long i = 0; i < (long long)((s).size()); ++i) {
    dp[i][1] = 1;
    for (long long j = 2; i - j + 1 >= 0 && i + j - 1 < (long long)((s).size());
         ++j) {
      if (s[i - j + 1] != s[i + j - 1]) break;
      dp[i - j + 1][j * 2 - 1] = 1;
    }
    if (i + 1 < (long long)((s).size()) && s[i] == s[i + 1]) {
      dp[i][2] = 2;
      for (long long j = 2; i - j + 1 >= 0 && i + j < (long long)((s).size());
           ++j) {
        if (s[i - j + 1] != s[i + j]) break;
        dp[i - j + 1][j * 2] = 1;
      }
    }
  }
  for (long long i = 2; i <= (long long)((s).size()); ++i) {
    for (long long l = 0; l + i <= (long long)((s).size()); ++l) {
      if (ok(l, i) && dp[l][i / 2] > 0)
        dp[l][i] = max(dp[l][i], dp[l][i / 2] + 1);
    }
  }
  for (long long i = 1; i <= (long long)((s).size()); ++i) {
    for (long long l = 0; l + i <= (long long)((s).size()); ++l) {
      ans[dp[l][i]]++;
    }
  }
  for (long long i = (long long)((s).size()) - 1; i > 0; --i)
    ans[i] += ans[i + 1];
  for (long long i = 1; i <= (long long)((s).size()); ++i)
    cout << ans[i] << ' ';
  return 0;
}