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;
long long mod = 1000000007;
long long z = 1000000000;
const long long N = 1e5 + 1;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
vector<vector<long long>> dp(5005, vector<long long>(5005, 0));
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    string s;
    cin >> s;
    long long n = s.length();
    for (long long l = 1; l <= n; l++) {
      for (long long i = 0; i < n - l + 1; i++) {
        long long j = i + l;
        if (l == 1) {
          dp[i][j] = 1;
          continue;
        } else if (l == 2) {
          if (s[i] == s[j - 1]) {
            dp[i][j] = 2;
          } else
            dp[i][j] = 0;
          continue;
        }
        if (s[i] != s[j - 1] || dp[i + 1][j - 1] == 0) continue;
        dp[i][j] = 1;
        long long m = (i + j) / 2;
        if (l & 1) {
          if (dp[i][m] & dp[m + 1][j]) dp[i][j] = dp[i][m] + 1;
        } else {
          if (dp[i][m] & dp[m][j]) dp[i][j] = dp[i][m] + 1;
        }
      }
    }
    vector<long long> cnt(n + 1, 0);
    for (int len = 1; len <= n; len++) {
      for (int l = 0; l < n - len + 1; l++) {
        cnt[dp[l][l + len]]++;
      }
    }
    for (long long i = n - 1; i >= 1; i--) {
      cnt[i] += cnt[i + 1];
    }
    for (long long i = 1; i < n + 1; i++) {
      cout << cnt[i] << " ";
    }
  }
}