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 = 5005;
char s[N];
int dp[N][N];
long long ans[N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) dp[i][i] = 1;
  for (int i = 1; i < n; i++)
    if (s[i] == s[i + 1]) dp[i][i + 1] = 2;
  for (int i = 3; i <= n; i++) {
    for (int j = 1; j + i - 1 <= n; j++) {
      if (s[j] == s[j + i - 1] && dp[j + 1][i + j - 2]) {
        dp[j][j + i - 1] = dp[j][j + i / 2 - 1] + 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; i + j - 1 <= n; j++) {
      ans[dp[j][j + i - 1]]++;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    ans[i] += ans[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    printf("%I64d%c", ans[i], i == n ? '\n' : ' ');
  }
}