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], ispal[5005][5005];
int cnt[5005], ans[5005];
int solve(int i, int j) {
  if (dp[i][j]) return dp[i][j];
  if (!ispal[i][j]) return dp[i][j] = 0;
  if (j == i) return dp[i][j] = 1;
  if (j == i + 1) return dp[i][j] = 2;
  int mid = (i + j) >> 1;
  if (!((j - i) & 1))
    return dp[i][j] = (min(solve(i, mid - 1), solve(mid + 1, j)) + 1);
  else
    return dp[i][j] = (min(solve(i, mid), solve(mid + 1, j)) + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  int n = s.size();
  s = "a" + s;
  for (int i = n; i >= 1; i--) {
    ispal[i][i] = 1;
    if (s[i] == s[i + 1]) ispal[i][i + 1] = 1;
    for (int j = i + 2; j <= n; j++)
      ispal[i][j] = (s[i] == s[j] && ispal[i + 1][j - 1]);
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) cnt[solve(i, j)]++;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++) ans[j] += cnt[i];
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}