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 n, a[1000005], ans[5006], dp[5006][5006];
char ch[1000005];
int main() {
  scanf("%s", ch);
  n = strlen(ch);
  for (int i = 0; i < n; i++) {
    dp[i][i] = 1;
    a[dp[i][i]]++;
    dp[i][i + 1] = (ch[i] == ch[i + 1]) * 2;
    a[dp[i][i + 1]]++;
  }
  for (int j = 2; j <= n; j++) {
    for (int i = 0; i + j <= n; i++) {
      int l = i;
      int r = i + j;
      if (ch[l] != ch[r] || dp[l + 1][r - 1] == 0) {
        continue;
      }
      int len = j + 1;
      int m = l + len / 2 - 1;
      dp[l][r] = 1;
      if (len & 1) {
        if (dp[l][m] && dp[m + 2][r]) {
          dp[l][r] = dp[l][m] + 1;
        }
      } else {
        if (dp[l][m] && dp[m + 1][r]) {
          dp[l][r] = dp[l][m] + 1;
        }
      }
      a[dp[l][r]]++;
    }
  }
  for (int i = n; i >= 0; i--) {
    ans[i] = ans[i + 1] + a[i];
  }
  for (int i = 1; i <= n; i++) {
    printf("%d%c", ans[i], i == n ? '\n' : ' ');
  }
  return 0;
}