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 = 5010;
int tree[maxn];
int lowbit(int x) { return x & (-x); }
void add(int x, int v) {
  for (; x > 0; x -= lowbit(x)) tree[x] += v;
}
int sum(int x) {
  int res = 0;
  for (; x < maxn; x += lowbit(x)) res += tree[x];
  return res;
}
char s[maxn];
int dp[maxn][maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> (s + 1);
  int n = strlen(s + 1);
  for (int len = 1; len <= n; ++len) {
    for (int st = 1; st + len - 1 <= n; ++st) {
      int ed = st + len - 1;
      if (len <= 2) {
        if (s[st] != s[ed]) continue;
        dp[st][ed] = len;
        add(dp[st][ed], 1);
      } else {
        if (dp[st + 1][ed - 1] == 0 || s[st] != s[ed]) continue;
        int mid = (st + ed) / 2;
        dp[st][ed] = dp[mid + 1][ed] + 1;
        add(dp[st][ed], 1);
      }
    }
  }
  for (int j = 1; j <= n; ++j) cout << sum(j) << " ";
  cout << endl;
  return 0;
}