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 maxn = 5000 + 10;
string s, t;
int dp[maxn][maxn];
int ans[maxn];
int main() {
  string s;
  cin >> s;
  int len = s.length();
  memset(dp, 0, sizeof(dp));
  memset(ans, 0, sizeof(ans));
  ans[1] += len;
  for (int l = 0; l < len; l++) {
    int i = 0;
    while (l - i >= 0 && l + i < len && s[l - i] == s[l + i]) {
      dp[l - i][l + i] = 1;
      i++;
    }
    i = 0;
    while (l - i >= 0 && l + i < len && s[l - i] == s[l + 1 + i]) {
      dp[l - i][l + i + 1] = 1;
      i++;
    }
  }
  for (int ll = 2; ll <= len; ll++) {
    for (int l = 0; l + ll - 1 <= len - 1; l++) {
      int r = l + ll - 1;
      int mid = (l + r - 1) / 2;
      if (dp[l][mid] >= 1 && dp[l][r] >= 1)
        dp[l][r] = max(dp[l][r], dp[l][mid] + 1);
      ans[dp[l][r]]++;
    }
  }
  for (int i = len - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= len; i++) cout << ans[i] << " ";
  return 0;
}