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 = 5005;
int dp[N][N], ispal[N][N], ans[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  {
    string s;
    cin >> s;
    int n = s.length();
    for (long long l = n - 1; l >= 0; l--)
      for (int r = l; r < n; ++r)
        ispal[l][r] = (l + 1 >= r || ispal[l + 1][r - 1]) && s[l] == s[r];
    for (long long l = n - 1; l >= 0; l--) {
      for (int r = l; r < n; ++r) {
        if (ispal[l][r]) {
          dp[l][r] =
              min(dp[l][(r + l + 1 >> 1) - 1], dp[(l + r >> 1) + 1][r]) + 1;
          ++ans[dp[l][r]];
        }
      }
    }
    for (long long i = n - 1 - 1; i >= 0; i--) ans[i] += ans[i + 1];
    for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  }
}