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 N = 5e3;
char s[N + 10];
int n, bo[N + 10][N + 10], ans[N + 10];
int getk(int l, int r) {
  if (!bo[l][r]) return 0;
  if (l == r) return 1;
  if (l + 1 == r) return 2;
  int len = (r - l + 1) / 2;
  int L = l, R = r;
  return getk(L, L + len - 1) + 1;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) {
    bo[i][i] = 1;
    if (i + 1 <= n && s[i] == s[i + 1]) bo[i][i + 1] = 1;
  }
  for (int i = 3; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      int r = j + i - 1;
      if (r > n) break;
      if (bo[j + 1][r - 1] && s[j] == s[r]) bo[j][r] = 1;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      ans[getk(i, j)]++;
    }
  for (int i = n - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}