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;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using pill = pair<int, ll>;
using vvi = vector<vi>;
ll gcd(ll a, ll b) {
  ll t;
  while ((t = a % b) != 0) {
    a = b;
    b = t;
  }
  return b;
}
ll fast_exp(ll base, ll n, ll m) {
  if (n == 0) return 1;
  ll t = fast_exp(base, n / 2, m);
  if (n % 2 == 0)
    return (t * t) % m;
  else
    return (((t * t) % m) * base) % m;
}
bool is_set(int i, ll mask) { return (mask >> i) & 1; }
int count_bits(ll mask) {
  int ans = 0;
  for (int i = 0; i < 64; ++i)
    if (is_set(i, mask)) ++ans;
  return ans;
}
int first_bit(ll mask) {
  int i = 0;
  while (i < 64)
    if (is_set(i++, mask)) return i - 1;
  return -1;
}
int const N = 5000;
int dp[N + 1][N + 1];
int cnt[15];
int main() {
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  int n = s.length();
  s = " " + s;
  for (int i = 1; i <= n; ++i) {
    dp[i][i] = 1;
    if (i < n and s[i] == s[i + 1]) dp[i][i + 1] = 2;
  }
  for (int l = 3; l <= n; ++l)
    for (int i = 1; i <= n - l + 1; ++i) {
      int j = i + l - 1;
      if (s[i] != s[j] or dp[i + 1][j - 1] == 0)
        dp[i][j] = 0;
      else
        dp[i][j] = dp[i][(j + i + 1) / 2 - 1] + 1;
    }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) cnt[dp[i][j]]++;
  for (int i = 13; i > 0; --i) cnt[i] += cnt[i + 1];
  for (int i = 1; i < min(14, n + 1); ++i) cout << cnt[i] << " ";
  for (int i = 14; i < n + 1; ++i) cout << "0 ";
  cout << "\n";
  return 0;
}