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 N = 5200;
const int mod = 1000000000 + 19;
const int mod2 = 1000000000 + 147;
string s;
int n;
long long h[N], pw = 997, p[N];
long long h2[N], pw2 = 29, p2[N];
int ans[N];
long long get(int i, int j) {
  if (i > j) return 0;
  return ((h[j] - (h[i - 1] * p[j - i + 1])) % mod + mod) % mod;
}
long long get2(int i, int j) {
  assert(i <= j);
  return (h2[j] - (h2[i - 1] * p2[j - i + 1]) % mod2 + mod2) % mod2;
}
int d[N][N];
bool is[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  n = (int)s.size();
  s = "#" + s;
  p[0] = 1;
  p2[0] = 1;
  for (int i = 1; i <= n; ++i) {
    h[i] = (h[i - 1] * pw + s[i]) % mod;
    h2[i] = (h2[i - 1] * pw2 + s[i]) % mod2;
    p[i] = (p[i - 1] * pw) % mod;
    p2[i] = (p2[i - 1] * pw2) % mod2;
  }
  for (int i = 1; i <= n; ++i) {
    d[i][i] = 1;
    is[i][i] = 1;
    ans[1]++;
    if (i + 1 <= n) {
      if (s[i] == s[i + 1])
        d[i][i + 1] = 2, is[i][i + 1] = 1;
      else
        is[i][i + 1] = d[i][i + 1] = 0;
      ans[d[i][i + 1]]++;
    }
  }
  for (int len = 3; len <= n; ++len) {
    for (int i = 1; i + len - 1 <= n; ++i) {
      int j = i + len - 1;
      is[i][j] = (s[i] == s[j] && is[i + 1][j - 1]);
      if (is[i][j] && get(i, i + len / 2 - 1) == get(j - len / 2 + 1, j) &&
          get2(i, i + len / 2 - 1) == get2(j - len / 2 + 1, j)) {
        d[i][j] = d[i][i + len / 2 - 1] + 1;
      } else {
        if (is[i][j])
          d[i][j] = 1;
        else
          d[i][j] = 0;
      }
      ans[d[i][j]]++;
    }
  }
  for (int i = n; i > 0; --i) {
    ans[i] += ans[i + 1];
  }
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  return 0;
}