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;
long long mod(long long n) { return n % (long long)1000000007; }
int main() {
  string s;
  cin >> s;
  long long n = s.size();
  int h[n][n];
  memset(h, 0, sizeof(h));
  int a[n][n];
  memset(a, 0, sizeof(a));
  for (long long i = 0; i < n; i = i + 1) {
    long long l = i, r = i;
    while (l >= 0 && r < n && s[l] == s[r]) {
      a[l][r] = 1;
      l--;
      r++;
    }
  }
  for (long long i = 1; i < n; i = i + 1) {
    if (s[i] == s[i - 1]) {
      long long l = i - 1, r = i;
      while (l >= 0 && r < n && s[l] == s[r]) {
        a[l][r] = 1;
        l--;
        r++;
      }
    }
  }
  long long dp[n + 1];
  memset(dp, 0, sizeof(dp));
  for (long long i = 0; i < n; i = i + 1) {
    for (long long j = i; j < n; j = j + 1) {
      if (a[i][j]) {
        if (j - i == 0) {
          h[i][j] = 1;
          dp[1]++;
        } else {
          long long m = i + (j - i) / 2;
          if ((j - i) % 2 == 0) m--;
          if (a[i][m]) {
            h[i][j] = 1 + h[i][m];
            dp[h[i][j]]++;
          } else {
            h[i][j] = 1;
            dp[1]++;
          }
        }
      }
    }
  }
  for (long long i = n - 1; i > 0; i--) {
    dp[i] = dp[i] + dp[i + 1];
  }
  for (long long i = 1; i < n + 1; i = i + 1) cout << dp[i] << " ";
}