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 main() {
  string s;
  cin >> s;
  int n = s.length();
  int dp[n][n];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) dp[i][j] = 0;
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == s[i + 1])
      dp[i][i + 1] = 2;
    else
      dp[i][i + 1] = 0;
  }
  int diff = 2;
  for (int k = diff; k <= n - 1; k++) {
    for (int i = 0; i + k < n; i++) {
      int j = i + k;
      if (s[i] != s[j])
        dp[i][j] = 0;
      else if (dp[i + 1][j - 1] == 0)
        dp[i][j] = 0;
      else
        dp[i][j] = dp[i][(i + j - 1) / 2] + 1;
    }
  }
  int ans[n + 1], tmp[n + 1];
  for (int i = 0; i <= n; i++) ans[i] = 0, tmp[i] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      tmp[dp[i][j]]++;
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans[i] += tmp[j];
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}