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;
bool dp[2][5009][5009];
long long int hashes[5009][5009];
int main() {
  string str;
  cin >> str;
  int n = str.size();
  for (int i = 0; i < n; i++) {
    hashes[i][i] = str[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      hashes[i][j] = (hashes[i][j - 1] * 251 + str[j]) % 1000000007;
    }
  }
  int ans = 0;
  for (int len = 1; len <= n; len++) {
    for (int i = 0; i < n; i++) {
      int j = i + len - 1;
      dp[0][i][j] = (str[i] == str[j]) && (len < 3 || dp[0][i + 1][j - 1]);
      ans += (dp[0][i][j]);
    }
  }
  cout << ans << " ";
  int parity = 0;
  int min_length = 2;
  for (int k = 2; k <= n; k++) {
    ans = 0;
    if (min_length > n) {
      while (k <= n) {
        cout << 0 << " ";
        k++;
      }
      break;
    }
    parity = (parity + 1) % 2;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        dp[parity][i][j] = false;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int left_i, right_i, left_r, right_r;
        left_i = i;
        right_r = j;
        right_i = i + (j - i - 1) / 2;
        left_r = right_i + 1 + ((j - i) % 2 == 0);
        dp[parity][i][j] = dp[(parity + 1) % 2][left_i][right_i] &&
                           dp[(parity + 1) % 2][left_r][right_r] &&
                           hashes[left_i][right_i] == hashes[left_r][right_r];
        ans += dp[parity][i][j];
      }
    }
    cout << ans << " ";
    min_length *= 2;
  }
  cout << endl;
}