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;
using ll = long long;
using pll = pair<ll, ll>;
using dd = long double;
namespace {
static bool constexpr dbg = 0;
ll constexpr N = 5.1e3;
ll n, d[N][N], res[N];
string s;
void init() {
  cin >> s;
  n = s.size();
  s = " " + s;
}
void solve() {
  for (ll len = 1; len <= n; ++len) {
    for (ll l = 1; l + len - 1 <= n; ++l) {
      ll r = l + len - 1;
      if (len == 1) {
        d[l][r] = 1;
        continue;
      } else if (len == 2) {
        d[l][r] = (s[l] == s[r]) ? 2 : 0;
        continue;
      }
      if (s[l] != s[r] || !d[l + 1][r - 1]) continue;
      d[l][r] = 1;
      ll m = (l + r) / 2;
      if (len & 1) {
        if (d[l][m - 1] && d[m + 1][r]) d[l][r] = 1 + d[l][m - 1];
      } else {
        if (d[l][m] && d[m + 1][r]) d[l][r] = 1 + d[l][m];
      }
    }
  }
  for (ll i = 1; i <= n; ++i)
    for (ll j = i; j <= n; ++j) ++res[d[i][j]];
  for (ll i = n - 1; i >= 1; --i) res[i] += res[i + 1];
  for (ll i = 1; i <= (ll)(n); ++i) cout << res[i] << ' ';
}
}  // namespace
int main() {
  if (!dbg) {
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  cout << fixed << setprecision(20);
  init();
  solve();
  return 0;
}