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;
int dp[5005][5005], ans[5005];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> s;
  int n = s.length();
  for (int i = 0; i < n; i++) {
    int L = i - 1, R = i + 2;
    dp[i][i + 1] = 1;
    ans[1]++;
    for (int j = 0; L >= 0 && R <= n; j++) {
      if (s[L] == s[R - 1])
        dp[L][R] = 1;
      else
        break;
      L--;
      R++;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int L = i, R = i + 2;
    for (int j = 0; L >= 0 && R <= n; j++) {
      if (s[L] == s[R - 1])
        dp[L][R] = 1;
      else
        break;
      L--;
      R++;
    }
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j + i <= n; j++) {
      if (dp[j][j + i] == 0) continue;
      dp[j][j + i] = min(dp[j][j + i / 2], dp[j + i - i / 2][j + i]) + 1;
      ans[dp[j][j + i]]++;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    ans[i] += ans[i + 1];
  }
  cout << ans[1];
  for (int i = 2; i <= n; i++) {
    cout << " " << ans[i];
  }
  cout << endl;
  return 0;
}