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 MAXN = 5000 + 10;
char ss[MAXN * 2], s[MAXN];
int p[MAXN];
int ans[MAXN];
int ok[MAXN][MAXN];
int dp[MAXN][MAXN];
int main() {
  scanf("%s", s);
  int len = strlen(s);
  ans[1] = len;
  for (int i = 0; i < len; i++) ok[i][i] = 1;
  for (int i = 0; i < len; i++) {
    int l = i, r = i + 1;
    while (l >= 0 && r < len && s[l] == s[r]) {
      ok[l][r] = 1;
      ans[1]++;
      l--;
      r++;
    }
    l = i - 1, r = i + 1;
    while (l >= 0 && r < len && s[l] == s[r]) {
      ok[l][r] = 1;
      ans[1]++;
      l--;
      r++;
    }
  }
  for (int i = len - 1; i >= 0; i--)
    for (int j = len - 1; j >= 0; j--)
      if (i == j)
        dp[i][j] = dp[i + 1][j + 1] + 1;
      else if (s[i] == s[j])
        dp[i][j] = dp[i + 1][j + 1] + 1;
  for (int l = 1; l < len; l++)
    for (int i = 0; i + l < len; i++) {
      int j = i + l;
      int ll = l % 2 ? (l + 1) / 2 : l / 2;
      int p = i + ll - 1, q = (l % 2 ? p + 1 : p + 2);
      if (ok[i][p] && ok[q][j] && ok[i][p] == ok[q][j] && dp[i][q] >= ll) {
        ok[i][j] = ok[q][j] + 1;
        for (int rk = 2; rk <= ok[i][j]; rk++) ans[rk]++;
      }
    }
  for (int i = 1; i <= len; i++) printf("%d ", ans[i]);
}