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 base = 1e9 + 13;
const int maxn = 5010;
int n;
string s;
int res[maxn];
long long hashs[maxn], hasht[maxn], powerhash[maxn];
int f[maxn][maxn];
void nhap() {
  getline(cin, s);
  n = s.size();
  for (int i = 1; i <= n; i++) {
    hashs[i] = (hashs[i - 1] * 26 + int(s[i - 1])) % base;
  }
  for (int i = 1; i <= n; i++) {
    hasht[i] = (hasht[i - 1] * 26 + int(s[n - i])) % base;
  }
  powerhash[0] = 1;
  for (int i = 1; i <= n; i++) powerhash[i] = (powerhash[i - 1] * 26) % base;
}
long long gethashs(int i, int j) {
  long long t =
      (hashs[j] - (hashs[i - 1] * powerhash[j - i + 1] % base) + base) % base;
  return (t);
}
long long gethasht(int i, int j) {
  long long t =
      (hasht[j] - (hasht[i - 1] * powerhash[j - i + 1] % base) + base) % base;
  return (t);
}
void xuli() {
  for (int j = 1; j <= n; j++) {
    for (int i = 1; i <= j; i++) {
      if (gethashs(i, j) == gethasht(n - j + 1, n - i + 1)) {
        int mid = (j - i + 1) >> 1;
        int r = i + mid - 1;
        f[i][j] = (f[i][r] + 1);
        res[f[i][j]]++;
      }
    }
  }
  for (int i = n - 1; i >= 1; i--) res[i] = res[i] + res[i + 1];
  for (int i = 1; i <= n; i++) {
    printf("%d ", res[i]);
  }
}
int main() {
  nhap();
  xuli();
}