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 = 5005;
int dp[N][N], cnt[N], ans[N];
int main() {
  string str;
  cin >> str;
  int i, j, l, r, len = (int)str.length();
  for (i = 1; i <= len; i++) {
    for (l = 0; l < len - i + 1; l++) {
      r = l + i;
      if (i > 2) {
        int m = ((l + r) / 2);
        if (str[l] == str[r - 1] && dp[l + 1][r - 1] != 0) {
          dp[l][r] = 1;
          if (i & 1)
            dp[l][r] = (dp[l][m] && dp[m + 1][r]) ? dp[l][m] + 1 : 1;
          else
            dp[l][r] = (dp[l][m] && dp[m][r]) ? dp[l][m] + 1 : 1;
        } else {
        }
      } else {
        if (i == 1) {
          dp[l][r] = 1;
        } else {
          if (str[l] == str[r - 1]) {
            dp[l][r] = 2;
          } else {
            dp[l][r] = 0;
          }
        }
      }
    }
  }
  for (i = 1; i <= len; i++) {
    for (j = 0; j < len - i + 1; j++) ans[dp[j][j + i]]++;
  }
  for (i = len - 1; i >= 1; i--) {
    ans[i] += ans[i + 1];
  }
  for (i = 1; i <= len; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}