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;
int mod = 1e9 + 7;
long long powe(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
void fun() {}
string s;
int check[5001][5001];
int dp[5001][5001];
int n;
int cnt[5001];
int ist(int left, int right) {
  if (left >= right) return 1;
  if (check[left][right] != -1) return check[left][right];
  int ans = 0;
  if (s[left] == s[right]) ans |= ist(left + 1, right - 1);
  return check[left][right] = ans;
}
int fun(int left, int right) {
  if (left == right) return dp[left][right] = 1;
  if (dp[left][right] != -1) return dp[left][right];
  int ans = ist(left, right);
  if (ans) {
    int num = (right - left) / 2;
    int l = left + num - 1;
    if ((right - left) % 2 != 0) l++;
    return dp[left][right] = fun(left, l) + 1;
  }
  return dp[left][right] = 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  fun();
  cin >> s;
  n = s.length();
  memset(check, -1, sizeof(check));
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i > j) continue;
      if (dp[i][j] == -1) {
        fun(i, j);
      }
      cnt[dp[i][j]]++;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    cnt[i] += cnt[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    cout << cnt[i] << " ";
  }
}