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 INF = 1e9 + 10;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 10;
const int SMAXN = 1e3 + 10;
const int P = 2;
const long long BINF = 1e18 + 10;
const long double EPS = 1e-10;
const long double PI = acos(-1);
inline void error() {
  cout << -1;
  exit(0);
}
string s;
int n, ans[MAXN], k;
vector<vector<bitset<15> > > dp;
int main() {
  cin >> s;
  n = s.size();
  dp.resize(n + 1);
  for (register int i(1); i <= n; i++) {
    dp[i].resize(n + 1);
  }
  s.insert(s.begin(), ' ');
  for (register int i(1); i <= n; i++) {
    dp[i][i].set(1);
    ans[1]++;
    if (i < n) {
      if (s[i] == s[i + 1]) {
        dp[i][i + 1].set(1);
        ans[1]++;
      }
    }
  }
  for (int len = 3; len <= n; len++) {
    for (int l = 1; l <= n - len + 1; l++) {
      int r = l + len - 1;
      if (s[l] == s[r] && dp[l + 1][r - 1].test(1)) {
        dp[l][r].set(1);
        ans[1]++;
      }
    }
  }
  int k = int(log2(n)) + 1;
  for (int i = 2; i <= k; i++) {
    for (int l = 1; l <= n; l++) {
      for (int r = l + 1; r <= n; r++) {
        if (!dp[l][r].test(1)) continue;
        int len = r - l + 1, r2 = l + len / 2 - 1, l2 = r - len / 2 + 1;
        if (dp[l][r2].test(i - 1) && dp[l2][r].test(i - 1)) {
          dp[l][r].set(i);
          ans[i]++;
        }
      }
    }
  }
  for (register int i(1); i <= n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}