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;
struct ii {
  int a, b;
  bool operator<(ii o) const { return make_tuple(a, b) < make_tuple(o.a, o.b); }
};
struct iii {
  int a, b, c;
  bool operator<(iii o) const {
    return make_tuple(a, b, c) < make_tuple(o.a, o.b, o.c);
  }
};
int TC, N, M, K, Q, x, y, z;
string str;
int dp[5005][5005], cnt[5005];
int f(int l, int r) {
  int &ret = dp[l][r];
  if (ret != -1) return ret;
  if (l == r) {
    ret = 1;
    cnt[ret]++;
  } else if (l + 1 == r) {
    if (str[l] == str[r])
      ret = 2;
    else
      ret = 0;
    cnt[ret]++;
  } else {
    if (f(l + 1, r - 1) && str[l] == str[r])
      ret = f(l, (l + r - 1) / 2) + 1;
    else
      ret = 0;
    cnt[ret]++;
  }
  return ret;
}
int main() {
  cin >> str;
  N = str.length();
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < N; i++) {
    for (int j = i; j < N; j++) {
      f(i, j);
    }
  }
  for (int i = 1; i <= N; i++) {
    int ans = 0;
    for (int j = i; j <= N; j++) {
      ans += cnt[j];
    }
    if (i > 1) printf(" ");
    printf("%d", ans);
  }
  puts("");
  return 0;
}