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 MAXN = 5179;
int dp[MAXN][MAXN], f[MAXN];
int main() {
  string s;
  cin >> s;
  int n = s.size();
  for (int len = 1; len <= n; len++) {
    for (int i = 0; i < n - len + 1; i++) {
      int j = i + len;
      if (len == 1)
        dp[i][j] = 1;
      else if (len == 2) {
        if (s[i] == s[j - 1]) {
          dp[i][j] = 2;
        } else {
          dp[i][j] = 0;
        }
      } else if (s[i] == s[j - 1] && dp[i + 1][j - 1]) {
        dp[i][j] = 1;
        int k = (i + j) / 2;
        if (len % 2 == 1) {
          if (dp[i][k] && dp[k + 1][j]) {
            dp[i][j] = dp[i][k] + 1;
          }
        } else {
          if (dp[i][k] && dp[k][j]) {
            dp[i][j] = dp[i][k] + 1;
          }
        }
      }
    }
  }
  for (int len = 1; len <= n; len++) {
    for (int i = 0; i < n - len + 1; i++) {
      f[dp[i][i + len]]++;
    }
  }
  for (int i = n - 1; i > 0; i--) {
    f[i] += f[i + 1];
  }
  for (int i = 0; i < n; i++) {
    cout << f[i + 1] << " ";
  }
  cout << endl;
  return 0;
}