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 long long Llinf = LLONG_MAX;
const int Inf = INT_MAX;
const int Maxn = 5e3 + 10;
const int Mod = 1e9 + 7;
char* _c = new char[Maxn];
string GetString() {
  scanf("%s", _c);
  return string(_c);
}
string s;
int n;
int dp[Maxn][Maxn];
int pal[Maxn][Maxn];
int ans[Maxn];
int Calc(int l, int r) {
  if (l == r)
    return dp[l][l] = 1;
  else if (l + 1 == r)
    return dp[l][r] = (s[l] == s[r]) ? 2 : 0;
  if (dp[l][r] != -1) return dp[l][r];
  if (pal[l][r])
    return dp[l][r] = 1 + Calc(l, (l + r) / 2 - ((r - l + 1) & 1));
  else
    return dp[l][r] = 0;
}
int main() {
  memset(dp, -1, sizeof(dp));
  memset(pal, -1, sizeof(pal));
  s = GetString();
  n = s.size();
  for (int l = 1; l <= n; l++)
    for (int i = 0; i + l - 1 < n; i++)
      pal[i][l + i - 1] =
          (s[i] == s[l + i - 1] && (l < 3 || pal[i + 1][l + i - 1 - 1]));
  for (int l = 1; l <= n; l++)
    for (int i = 0; i + l - 1 < n; i++) ans[Calc(i, l + i - 1)]++;
  for (int i = n - 1; i >= 0; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}