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 = 5e3 + 10;
string s;
int best, n, p;
int l[MAXN], suff[MAXN], out[MAXN][26], dep[MAXN], half[MAXN], times[MAXN];
int ans[MAXN];
int getSuff(int v) {
  while (s[p] != s[p - 1 - l[v]]) v = suff[v];
  return v;
}
void addLetter(int v) {
  int cur = best;
  p = v;
  int let = s[v] - 'a';
  cur = getSuff(cur);
  if (!out[cur][let]) {
    n++;
    suff[n] = out[getSuff(suff[cur])][let];
    out[cur][let] = n;
    l[n] = l[cur] + 2;
    int ha = suff[n];
    while (l[ha] > l[n] / 2) ha = suff[ha];
    if (l[ha] > 0 && l[ha] == l[n] / 2)
      dep[n] = dep[ha] + 1;
    else
      dep[n] = 1;
    half[n] = ha;
  }
  int o = out[cur][let];
  best = o;
  times[o]++;
}
int main() {
  l[1] = -1, suff[1] = 1, dep[1] = 0;
  l[0] = 0, suff[0] = 1, dep[0] = 0;
  best = 1, n = 1;
  cin >> s;
  for (int i = 0; i < s.size(); i++) addLetter(i);
  for (int i = n; i > 1; i--) {
    times[suff[i]] += times[i];
    ans[dep[i]] += times[i];
  }
  int cur = 0;
  for (int i = s.size(); i > 0; i--) {
    cur += ans[i];
    ans[i] = cur;
  }
  for (int i = 1; i <= s.size(); i++) printf("%d ", ans[i]);
}