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;
int dp[5001][5001], ans[5100], s;
char t[5100];
int main() {
  scanf("%s", t);
  s = strlen(t);
  for (int i = 0; i < s; i++) {
    dp[i][i] = 1;
    if (i > 0 && t[i] == t[i - 1]) dp[i - 1][i] = 2;
  }
  for (int i = 2; i < s; i++)
    for (int j = 0; j + i < s; j++)
      if (t[j] == t[j + i] && dp[j + 1][j + i - 1]) {
        int m;
        if (i % 2)
          m = i / 2, dp[j][j + i] = dp[j][j + m] + 1;
        else
          m = -1 + i / 2, dp[j][j + i] = dp[j][j + m] + 1;
      }
  for (int i = 0; i < s; i++)
    for (int j = i; j < s; j++) ans[dp[i][j]]++;
  for (int i = s - 1; i >= 0; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= s; i++) printf("%d ", ans[i]);
  printf("\n");
}