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>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
#pragma GCC optimize("Ofast")
using namespace std;
using lint = long long;
int16_t get_k[5555][5555];
int ans[5555];
int16_t lcp[5555][5555];
char buf[int(1e5)];
size_t bp = 0;
void pri(int x) {
  int len = 0;
  do {
    buf[bp + len] = char('0' + x % 10);
    x /= 10;
    len++;
  } while (x);
  reverse(buf + bp, buf + bp + len);
  bp += len;
}
void solve(istream& cin, ostream& cout) {
  string s;
  cin >> s;
  const int n = s.size();
  for (int i = n - 1; i >= 0; i--) {
    for (int j = n - 1; j >= 0; j--) {
      if (s[i] == s[j]) {
        lcp[i][j] = lcp[i + 1][j + 1] + 1;
      }
    }
  }
  for (int i = (0); i < int(n); ++i) {
    get_k[i][i] = true;
  }
  ans[0] = n;
  for (int len = 2; len <= n; len++) {
    for (int l = 0; l + len <= n; l++) {
      int r = l + len - 1;
      if (s[l] == s[r] && (l + 1 == r || get_k[l + 1][r - 1])) {
        if (l + 1 == r) {
          get_k[l][r] = 2;
          ans[1]++;
        } else {
          int m = len >> 1;
          if (lcp[l][r - m + 1] >= m) {
            int x = get_k[l][l + m - 1];
            get_k[l][r] = x + 1;
            ans[x]++;
          } else {
            get_k[l][r] = 1;
            ans[0]++;
          }
        }
      }
    }
  }
  for (int i = int(n - 1) - 1; i >= (0); --i) {
    ans[i] += ans[i + 1];
  }
  for (int i = (0); i < int(n); ++i) {
    pri(ans[i]);
    buf[bp++] = ' ';
  }
  puts(buf);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve(cin, cout);
  return 0;
}