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 long long int N = 5001;
const long long int inf = 1ll << 62;
using pii = pair<long long int, long long int>;
const long long int mod = 1e9 + 7;
long long int powm(long long int a, long long int b) {
  a = a % mod;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int dp[N][N];
int is_palindrome[N][N];
string s;
int rec1(long long int i, long long int j) {
  if (is_palindrome[i][j] != -1) {
    return is_palindrome[i][j];
  }
  if (i > j) {
    return 0;
  }
  if (i == j) {
    is_palindrome[i][j] = 1;
    return is_palindrome[i][j];
  }
  if (i + 1 == j) {
    is_palindrome[i][j] = s[i] == s[j];
    return is_palindrome[i][j];
  } else {
    is_palindrome[i][j] = 0;
    if (s[i] == s[j] && rec1(i + 1, j - 1)) {
      is_palindrome[i][j] = 1;
    }
    return is_palindrome[i][j];
  }
}
int rec(long long int i, long long int j) {
  if (dp[i][j] != -1) {
    return dp[i][j];
  }
  if (i == j) {
    dp[i][j] = 1;
    return dp[i][j];
  } else {
    long long int len = (j - i + 1) / 2;
    dp[i][j] = 0;
    if (rec1(i, j)) {
      dp[i][j] = min(rec(i, i + len - 1), rec(j - len + 1, j)) + 1;
    }
    return dp[i][j];
  }
}
int main() {
  long long int t = 1;
  memset(dp, -1, sizeof(dp));
  memset(is_palindrome, -1, sizeof(is_palindrome));
  cin >> s;
  long long int n = s.size();
  long long int ans[n + 2];
  memset(ans, 0, sizeof(ans));
  for (int i = 0; i < s.size(); i++) {
    for (int j = i; j < s.size(); j++) {
      long long int a = rec(i, j);
      ans[a + 1]--;
      ans[0]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    ans[i] += ans[i - 1];
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
  return 0;
}