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;
struct pair_hash {
  template <typename T1, typename T2>
  size_t operator()(const pair<T1, T2> &p) const {
    auto h1 = hash<T1>{}(p.first);
    auto h2 = hash<T2>{}(p.second);
    return h1 ^ h2;
  }
};
const long long INFLL = 2 * (long long)1e18 + 100;
const int INFINT = 2 * (int)1e9 + 100;
void die() {
  cout << "NO";
  exit(0);
}
const int NMAX = 5 * 1e3 + 5;
const int MOD = 1e9 + 7;
const double PI = atan(1) * 4;
const double EPS = 1e-12;
string s;
int n;
char v[NMAX];
bool is_pal[NMAX][NMAX];
int dp[NMAX][NMAX], sol[NMAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  for (int i = 0; i < s.size(); i++) v[i + 1] = s[i];
  n = s.size();
  for (int i = 1; i <= n; i++) {
    is_pal[i][i] = 1;
    dp[i][i] = 1;
    sol[1]++;
  }
  for (int len = 2; len <= n; len++) {
    for (int i = 1; i <= n; i++) {
      if (len + i - 1 > n) break;
      if (v[i] == v[len + i - 1] && (is_pal[i + 1][len + i - 2] || len == 2)) {
        is_pal[i][i + len - 1] = 1;
      }
    }
  }
  for (int len = 2; len <= n; len++) {
    for (int i = 1; i <= n; i++) {
      if (len + i - 1 > n) break;
      if (is_pal[i][i + len - 1] && is_pal[i][i + len / 2 - 1] &&
          is_pal[i + len / 2 + len % 2][i + len - 1]) {
        dp[i][i + len - 1] = dp[i][i + len / 2 - 1] + 1;
        sol[dp[i][i + len - 1]]++;
      } else if (is_pal[i][i + len - 1]) {
        dp[i][i + len - 1] = 1;
        sol[1]++;
      }
    }
  }
  for (int i = n; i >= 1; i--) sol[i] += sol[i + 1];
  for (int i = 1; i <= n; i++) cout << sol[i] << ' ';
  return 0;
}