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;
char str[5010];
int dp[5010][5010], ans[5010];
int main() {
  scanf("%s", str);
  int len = strlen(str);
  for (int i = 0; i < len; i++) {
    dp[i][i] = 1;
    ans[1]++;
    for (int j = i - 1, k = i + 1; j >= 0 && k < len; j--, k++) {
      if (str[j] != str[k]) break;
      dp[j][k] = 1;
      ans[1]++;
    }
    for (int j = i, k = i + 1; j >= 0 && k < len; j--, k++) {
      if (str[j] != str[k]) break;
      dp[j][k] = 1;
      ans[1]++;
    }
  }
  for (int k = 2; k <= len; k++) {
    for (int i = 0; i < len; i++) {
      int l = i + k - 1;
      if (l > len) break;
      int mid = i + k / 2 - 1;
      if (dp[i][l] && dp[i][mid]) {
        dp[i][l] = max(dp[i][l], dp[i][mid] + 1);
        ans[dp[i][l]]++;
      }
    }
  }
  for (int i = len - 1; i >= 2; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= len; i++) printf("%d ", ans[i]);
  return 0;
}