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;
string a;
int dp[5010][5010];
int cnt[100010];
int main() {
  cin >> a;
  int t = a.size();
  for (int i = 0; i < t; i++) {
    dp[i][i] = 1;
    cnt[1]++;
  }
  for (int i = 0; i < t - 1; i++) {
    if (a[i] == a[i + 1]) {
      dp[i][i + 1] = 2;
      cnt[2]++;
    }
  }
  for (int i = 2; i < t; i++) {
    for (int y = 0; y < t; y++) {
      if (i + y >= t) continue;
      if (a[y] == a[i + y] && dp[y + 1][i + y - 1]) {
        int mid = ((i + 1) / 2 - 1);
        int ch = min(dp[y][mid + y], dp[i + y - mid][i + y]);
        dp[y][y + i] = ch + 1;
        cnt[ch + 1]++;
      }
    }
  }
  for (int i = 100000; i >= 0; i--) cnt[i] += cnt[i + 1];
  for (int i = 1; i <= t; i++) printf("%d ", cnt[i]);
}