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 mod = 1e9 + 7;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int main() {
  string second;
  cin >> second;
  int n = second.size();
  int dp[n + 1][n + 1];
  memset(dp, 0, sizeof(dp));
  int seen[n + 1][n + 1];
  int ans[n + 1];
  memset(ans, 0, sizeof(ans));
  for (int len = 1; len <= n; len++) {
    for (int i = 0, j = i + len - 1; j < n; i++, j++) {
      if (len == 1)
        dp[i][j] = 1;
      else if (second[i] != second[j])
        continue;
      else if (len == 2)
        dp[i][j] = 2;
      else if (dp[i + 1][j - 1] == 0)
        continue;
      else {
        int mid = (i + j) >> 1;
        if (len & 1)
          dp[i][j] = min(dp[i][mid - 1], dp[mid + 1][j]) + 1;
        else
          dp[i][j] = min(dp[i][mid], dp[mid + 1][j]) + 1;
      }
      ans[dp[i][j]]++;
    }
  }
  for (int i = n - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}