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 MOD = 998244353, maxN = 5e3 + 6, maxV = 6561 + 5;
string st;
int maxPal[maxN][maxN], n, ans[maxN], dp[maxN][maxN], acc[maxN];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> st;
  n = int(st.size());
  for (int i = 0; i < n; i++) {
    int k = 0;
    while (i - k >= 0 && i + k < n && st[i - k] == st[i + k]) {
      maxPal[i - k][k * 2 + 1] = 1;
      k++;
    }
  }
  for (int i = 0; i < n; i++) {
    int k = 1;
    while (i - k >= 0 && i + k - 1 < n && st[i - k] == st[i + k - 1]) {
      maxPal[i - k][k * 2] = 1;
      k++;
    }
  }
  memcpy(dp, maxPal, sizeof(dp));
  for (int u = 1; u <= n; u++) {
    for (int i = 0; i + u <= n; i++) {
      if (maxPal[i][u]) {
        dp[i][u] = dp[i][u >> 1] + 1;
        acc[1]++, acc[dp[i][u] + 1]--;
      }
    }
  }
  int ct = 0;
  for (int i = 1; i <= n; i++) {
    ct += acc[i];
    cout << ans[i] + ct << ' ';
  }
  cout << '\n';
}