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], res[5001];
char s[5001];
int main() {
  scanf("%s", s);
  int dl = strlen(s);
  for (int i = 0; i < dl; ++i) res[i] = 0;
  for (int i = 0; i < dl - 1; ++i) {
    dp[i][i] = 1;
    if (s[i] == s[i + 1])
      dp[i][i + 1] = 2;
    else
      dp[i][i + 1] = 0;
  }
  dp[dl - 1][dl - 1] = 1;
  for (int i = 2; i < dl; ++i)
    for (int l = 0; l + i < dl; ++l) {
      if (s[l] == s[l + i] && dp[l + 1][l + i - 1] > 0)
        dp[l][l + i] = dp[l][(2 * l + i - 1) / 2] + 1;
      else
        dp[l][l + i] = 0;
    }
  for (int i = 0; i < dl; ++i)
    for (int j = i; j < dl; ++j) res[dp[i][j]]++;
  for (int i = dl - 1; i > 0; --i) res[i] += res[i + 1];
  for (int i = 1; i <= dl; ++i) printf("%d ", res[i]);
}