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 maxn = 5010;
const unsigned long long mul = 1e9 + 7;
string s, r;
int res[maxn], sum[maxn][maxn];
unsigned long long hash1[maxn], hash2[maxn], p[maxn];
bool palindrome(int l, int r) {
  unsigned long long h1 = hash1[r + 1] - hash1[l] * p[r - l + 1];
  int l2 = s.length() - 1 - r, r2 = s.length() - 1 - l;
  unsigned long long h2 = hash2[r2 + 1] - hash2[l2] * p[r2 - l2 + 1];
  return h1 == h2;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> s;
  r = s;
  reverse(r.begin(), r.end());
  p[0] = 1;
  for (int i = 1; i <= s.length(); i++) {
    p[i] = p[i - 1] * mul;
    hash1[i] = hash1[i - 1] * mul + s[i - 1];
    hash2[i] = hash2[i - 1] * mul + r[i - 1];
  }
  for (int i = 0; i < s.length(); i++) {
    sum[i][i] = 1;
    res[1]++;
  }
  for (int len = 2; len <= s.length(); len++) {
    for (int i = 0; i + len - 1 < s.length(); i++) {
      int j = i + len - 1;
      if (palindrome(i, j)) {
        sum[i][j] = sum[i][i + (j - i + 1) / 2 - 1] + 1;
        res[sum[i][j]]++;
      }
    }
  }
  for (int i = s.length(); i >= 1; i--) {
    res[i] += res[i + 1];
  }
  for (int i = 1; i <= s.length(); i++) {
    cout << res[i] << " ";
  }
  cout << endl;
  return 0;
}