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;
int dp[5005][5005];
int ans[5005];
int out[5005];
int main() {
  ios::sync_with_stdio(0);
  string s;
  cin >> s;
  int N = (int)s.size();
  for (int k = 0; k < N; ++k) {
    for (int i = 0; i < N; ++i) {
      int j = i + k;
      if (j >= N) break;
      if (k / 2 == 0) {
        if (s[i] == s[j]) dp[i][j] = k + 1;
        ++ans[dp[i][j]];
      } else {
        if (s[i] == s[j]) {
          if (dp[i + 1][j - 1]) {
            dp[i][j] = dp[i][i + (k + 1) / 2 - 1] + 1;
            ++ans[dp[i][j]];
          }
        }
      }
    }
  }
  for (int i = N - 1; i >= 1; --i) {
    out[i] = out[i + 1] + ans[i];
  }
  for (int i = 1; i <= N; ++i) {
    cout << out[i] << ' ';
  }
  cout << '\n';
}