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 = 1e5 + 10;
const int MAXM = 1e6 + 10;
const int BASE = 1e9 + 7;
bool pal[5001][5001];
int f[5001][5001];
class SOLUTION {
 public:
  string st;
  int n;
  int cnt[5001];
  void read_input() {
    cin >> st;
    n = st.size();
  }
  void init() {
    memset(pal, false, sizeof(pal));
    for (int i = 1; i <= n; i++) {
      pal[i][i] = true;
      if (i < n && st[i - 1] == st[i]) {
        pal[i][i + 1] = true;
      }
    }
    for (int i = 2; i < n; i++) {
      for (int j = 1; j <= n - i; j++) {
        if (st[j - 1] == st[i + j - 1]) {
          pal[j][i + j] = pal[j + 1][i + j - 1];
        }
      }
    }
  }
  void solve() {
    for (int i = 1; i <= n; i++) {
      f[i][i] = 1;
    }
    for (int i = 1; i < n; i++) {
      for (int j = 1; j <= n - i; j++) {
        if (pal[j][i + j]) {
          f[j][i + j] = 1;
          int len = (i + 1) / 2;
          if (pal[j][j + len - 1] && pal[i + j - len + 1][i + j]) {
            f[j][i + j] = min(f[j][j + len - 1], f[i + j - len + 1][i + j]) + 1;
          }
        }
      }
    }
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) {
      for (int j = i; j <= n; j++) {
        if (pal[i][j]) {
          cnt[f[i][j]]++;
        }
      }
    }
    for (int i = n - 1; i >= 1; i--) {
      cnt[i] += cnt[i + 1];
    }
    for (int i = 1; i < n; i++) {
      cout << cnt[i] << " ";
    }
    cout << cnt[n] << endl;
  }
  void execute() {
    int test = 1;
    while (test--) {
      read_input();
      init();
      solve();
    }
  }
};
int main() {
  SOLUTION sol;
  sol.execute();
  return 0;
}