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;
string s;
int pal[5010][5010], dp[5010][5010], ans[5010];
int ispal(int x, int y) {
  if (x > y || x == y) return 1;
  int &ret = pal[x][y];
  if (ret != -1) return ret;
  ret = 0;
  if (s[x] == s[y]) ret = ispal(x + 1, y - 1);
  pal[x][y] = ret;
  return ret;
}
int fun(int x, int y) {
  if (x == y) return 1;
  int &ret = dp[x][y];
  if (ret != -1) return ret;
  ret = 0;
  if (ispal(x, y)) ret = 1 + fun(x, (x + y - 1) / 2);
  dp[x][y] = ret;
  return ret;
}
int main() {
  int n;
  cin >> s;
  n = s.size();
  memset(dp, -1, sizeof(dp));
  memset(pal, -1, sizeof(pal));
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      int x = fun(i, j);
      ans[x + 1]--;
      ans[1]++;
    }
  }
  for (int i = 1; i <= n; i++) ans[i] += ans[i - 1];
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
}