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 MAXN = 5e3 + 5;
int n;
string s;
int f(char c) { return c - 'a' + 1; }
const long long MOD = 1e9 + 7;
const long long base = 37;
int dp[MAXN][MAXN];
int maxPalindrome(int l, int r) {
  if (l == r) return 1;
  if (r - l == 1) return (s[l] == s[r] ? 2 : 0);
  if (dp[l][r] != -1) return dp[l][r];
  if (s[l] != s[r] || maxPalindrome(l + 1, r - 1) == 0) return dp[l][r] = 0;
  int mid = (l + r) / 2;
  int left = maxPalindrome(l, mid - ((r - l + 1) % 2 == 1));
  int right = maxPalindrome(mid + 1, r);
  int minMax = min(left, right);
  return dp[l][r] = minMax + 1;
}
int mpFreq[MAXN];
int ans[MAXN];
void solve() {
  cin >> s;
  n = s.length();
  s = ' ' + s;
  memset(dp, -1, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      int k = maxPalindrome(i, j);
      mpFreq[k]++;
    }
  }
  long long lastSum = 0;
  for (int i = n; i >= 0; i--) {
    long long val = mpFreq[i] + lastSum;
    ans[i] = val;
    lastSum += mpFreq[i];
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  solve();
}