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 INF = 1e9;
const long long MAX = 5e3 + 5;
const long long MOD = 1e9 + 7;
int sz;
string s;
int arr[MAX][MAX];
long long has[MAX];
long long pp[MAX];
bool cmp(int i, int j, int l) {
  long long a = has[i + l - 1] - (i - 1 >= 0 ? has[i - 1] : 0);
  long long b = has[j + l - 1] - (j - 1 >= 0 ? has[j - 1] : 0);
  a = (a + MOD) % MOD;
  b = (b + MOD) % MOD;
  return a * pp[j - i] % MOD == b;
}
void build() {
  pp[0] = 1;
  for (int i = 1; i <= sz; i++) {
    pp[i] = pp[i - 1] * 26;
    pp[i] %= MOD;
  }
  has[0] = s[0] - '0';
  for (int i = 1; i < sz; i++) {
    has[i] = has[i - 1] + (s[i] - '0') * pp[i] % MOD;
    has[i] %= MOD;
  }
  for (int l = 3; l <= sz; l++)
    for (int i = 0; i + l <= sz; i++)
      if (arr[i + 1][l - 2] && s[i] == s[i + l - 1]) arr[i][l] = 1;
}
void pro() {
  for (int l = 2; l <= sz; l++) {
    int hl = l / 2;
    for (int i = 0; i + l <= sz; i++)
      if (arr[i][hl] && arr[i + l - hl][hl] && cmp(i, i + l - hl, hl))
        arr[i][l] = min(arr[i][hl], arr[i + l - hl][hl]) + 1;
  }
  return;
}
int cnt[MAX];
int main() {
  do {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  } while (0);
  cin >> s;
  sz = ((long long)(s).size());
  for (int i = 0; i < sz; i++) arr[i][1] = 1;
  for (int i = 0; i < sz - 1; i++)
    if (s[i] == s[i + 1]) arr[i][2] = 1;
  build();
  pro();
  for (int i = 0; i < sz; i++) {
    for (int j = 1; j <= sz; j++) {
      cnt[arr[i][j]]++;
    }
  }
  for (int i = 12; i > 0; i--) cnt[i] += cnt[i + 1];
  for (int i = 1; i <= sz; i++) {
    if (i != 1) cout << ' ';
    cout << cnt[i];
  }
  cout << '\n';
  return 0;
}