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 N = 5000 + 11;
unsigned long long dp1[N][N];
bool ans1[N][N], ans2[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    unsigned long long d = 0;
    unsigned long long p = 1;
    for (int j = i; j < s.size(); j++) {
      d += p * (s[j] - 'a' + 1);
      p *= 29;
      dp1[i + 1][j + 1] = d;
    }
  }
  int ans = 0;
  int n = s.size();
  reverse(s.begin(), s.end());
  for (int i = 0; i < s.size(); i++) {
    unsigned long long d = 0;
    unsigned long long p = 1;
    for (int j = i; j < s.size(); j++) {
      d += p * (s[j] - 'a' + 1);
      p *= 29;
      if (dp1[n - j][n - i] == d) {
        ans++;
        ans1[n - j][n - i] = 1;
      }
    }
  }
  cout << ans << " ";
  for (int k = 2; k <= min(n, 13); k++) {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) {
        ans2[i][j] = ans1[i][j];
        ans1[i][j] = 0;
      }
    ans = 0;
    for (int i = 1; i <= n; i++)
      for (int j = i + 1; j <= n; j++) {
        int sz = (j - i + 1) / 2;
        if (dp1[i][i + sz - 1] == dp1[j - sz + 1][j] &&
            ans2[i][i + sz - 1] == 1) {
          ans1[i][j] = 1;
          ans++;
        }
      }
    cout << ans << " ";
  }
  for (int i = 14; i <= n; i++) cout << 0 << " ";
}