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 memo[5004][5004], maxk[5004][5004] = {0};
int cnt[5004] = {0};
int main() {
  char str[5004];
  scanf("%s", str);
  int n = strlen(str);
  for (int i = 0; i < n; i++) {
    long long int h = 0, rh = 0, p = 47, pw = 1;
    for (int j = i; j < n; j++) {
      int v = str[j] - 'a' + 1;
      h = (h * p + v) % 1000000007;
      rh = (rh + v * pw) % 1000000007;
      pw = (pw * p) % 1000000007;
      memo[i][j] = (int)h;
      memo[j][i] = (int)rh;
    }
  }
  for (int l = 1; l <= n; l++) {
    for (int i = 0, j = i + l - 1; j < n; i++, j++) {
      if (memo[i][j] == memo[j][i]) {
        maxk[i][j] = 1;
        if (l & 1) {
          int m = (i + j) / 2;
          if (memo[i][m - 1] == memo[m + 1][j]) {
            maxk[i][j] = max(maxk[i][j], maxk[m + 1][j] + 1);
          }
        } else {
          int m = (i + j) / 2;
          if (memo[i][m] == memo[m + 1][j]) {
            maxk[i][j] = max(maxk[i][j], maxk[m + 1][j] + 1);
          }
        }
        cnt[maxk[i][j]]++;
      }
    }
  }
  for (int i = n - 1; i > 0; i--) {
    cnt[i - 1] += cnt[i];
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", cnt[i]);
  }
  printf("\n");
  return 0;
}