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>
template <class T, class U>
inline bool smin(T &x, const U &y) {
  return y < x ? x = y, 1 : 0;
}
template <class T, class U>
inline bool smax(T &x, const U &y) {
  return x < y ? x = y, 1 : 0;
}
using LL = long long;
using PII = std::pair<int, int>;
constexpr int N(5005);
std::array<int, 26> ch[N];
int fa[N], len[N], cnt, f[N], val[N], g[N];
char s[N], *now;
int jmp(int x) {
  for (; now[~len[x]] != *now; x = fa[x])
    ;
  return x;
}
int ins(int x) {
  static int p, np;
  if (!ch[p = jmp(p)][x]) {
    len[np = ++cnt] = len[p] + 2;
    fa[np] = ch[jmp(fa[p])][x];
    ch[p][x] = np;
    if (len[np] <= 2) {
      f[np] = fa[np];
    } else {
      for (int &i = f[np] = f[p];
           now[~len[i]] != *now || (len[i] + 2 << 1) > len[np]; i = fa[i])
        ;
      f[np] = ch[f[np]][x];
    }
  }
  p = ch[p][x];
  val[p]++;
  return p;
}
int ans[N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  len[1] = -1;
  fa[0] = 1;
  cnt = 1;
  std::cin >> s + 1;
  now = s;
  int n = strlen(s + 1);
  for (int i = 1; i <= n; i++) {
    ins(*++now - 'a');
  }
  for (int i = cnt; i > 1; i--) val[fa[i]] += val[i];
  for (int i = 2; i <= cnt; i++) {
    if (len[i] / 2 == len[f[i]]) {
      g[i] = g[f[i]] + 1;
    } else {
      g[i] = 1;
    }
    ans[g[i]] += val[i];
  }
  for (int i = n - 1; i; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) std::cout << ans[i] << " ";
  return 0;
}