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 dp[5050][5050], q[5050][5050];
int ans[5050], len;
char s[5050];
void init() {
  for (int i = 0; i < len; i++) {
    int l = 0;
    while (i - l >= 0 && i + l < len && s[i - l] == s[i + l])
      dp[i - l][i + l] = 1, l++;
    l = 0;
    while (i - l >= 0 && i + 1 + l < len && s[i - l] == s[i + l + 1])
      dp[i - l][i + l + 1] = 1, l++;
  }
}
void solve() {
  for (int t = 0; t < len; t++) {
    for (int j = 0; j + t < len; j++) {
      if (!dp[j][j + t]) continue;
      int mid;
      if (t % 2 == 0)
        mid = (j + j + t) / 2 - 1;
      else
        mid = (j + j + t) / 2;
      if (mid < j)
        q[j][j + t] = 1;
      else
        q[j][j + t] = q[j][mid] + 1;
      ans[q[j][j + t]]++;
    }
  }
}
int main() {
  scanf("%s", s);
  len = strlen(s);
  memset(dp, 0, sizeof(dp)), memset(ans, 0, sizeof(ans)),
      memset(q, 0, sizeof(q));
  init();
  solve();
  for (int i = len - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= len; i++) {
    if (i == 1)
      printf("%d", ans[i]);
    else
      printf(" %d", ans[i]);
  }
  printf("\n");
  return 0;
}