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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void solve() {
  string s;
  cin >> s;
  int n = s.size();
  int dp[n][n];
  memset(dp, 0, sizeof dp);
  for (int i = 0; i < n; ++i) {
    dp[i][i] = 1;
  }
  for (int i = 0; i < n - 1; ++i) {
    if (s[i] == s[i + 1]) dp[i][i + 1] = 1;
  }
  for (int len = 3; len <= n; ++len) {
    for (int i = 0; i < n - len + 1; ++i) {
      int j = i + len - 1;
      if (s[i] == s[j]) dp[i][j] = dp[i + 1][j - 1];
    }
  }
  int cur[n][n];
  memset(cur, 0, sizeof cur);
  for (int i = 0; i < n; ++i) {
    cur[i][i] = 1;
    if (i + 1 < n && s[i] == s[i + 1]) cur[i][i + 1] = 2;
  }
  for (int len = 3; len <= n; ++len) {
    for (int i = 0; i < n - len + 1; ++i) {
      int j = i + len - 1;
      if (len & 1) {
        if (dp[i][j] && cur[i][i + len / 2 - 1] == cur[i + len / 2 + 1][j])
          cur[i][j] = cur[i][i + len / 2 - 1] + 1;
      } else {
        if (dp[i][j] && cur[i][i + len / 2 - 1] == cur[i + len / 2][j])
          cur[i][j] = cur[i][i + len / 2 - 1] + 1;
      }
      cur[i][j] = max(dp[i][j], cur[i][j]);
    }
  }
  vector<int> ans(n + 1);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      ans[cur[i][j]]++;
    }
  }
  for (int i = n - 1; i >= 0; --i) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  while (t--) solve();
}