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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
const int N = 5005;
int dp[N][N], ans[N];
bool ok[N][N];
char s[N];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  ans[1] = n;
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int i = 0; i < n; i++) {
    int st = i, en = i;
    while (st >= 0 and en < n and s[st] == s[en]) {
      ok[st][en] = 1;
      st--;
      en++;
    }
    st = i;
    en = i + 1;
    while (st >= 0 and en < n and s[st] == s[en]) {
      ok[st][en] = 1;
      st--;
      en++;
    }
  }
  for (int len = 2; len <= n; len++)
    for (int l = 0; l + len <= n; l++) {
      int r = l + len - 1;
      if (ok[l][r]) dp[l][r] = dp[l][l + len / 2 - 1] + 1;
      ans[dp[l][r]]++;
    }
  for (int i = n; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}