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 n;
char s[5005];
int lvls[5002][5002];
bool vis[5002][5002];
int ans[5005];
inline void init() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
}
inline int getLvl(int l, int r) {
  if (l == r) return 1;
  if (lvls[l][r] != 1 || vis[l][r]) return lvls[l][r];
  vis[l][r] = true;
  int len = (r - l + 1) / 2;
  return lvls[l][r] = (getLvl(l, l + len - 1) + 1);
}
inline void solve() {
  s[0] = '+', s[n + 1] = '-', s[n + 2] = 0;
  for (int i = 1; i <= n; i++) {
    lvls[i][i] = 1;
    int l = i - 1, r = i + 1;
    while (s[l] == s[r]) lvls[l][r] = 1, l--, r++;
    l = i, r = i + 1;
    while (s[l] == s[r]) lvls[l][r] = 1, l--, r++;
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      ans[getLvl(i, j)]++;
    }
  for (int i = n - 1; i; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
int main() {
  init();
  solve();
  return 0;
}