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;
namespace MY {
const int mod = 10000007;
const int maxn = 5005;
char s[maxn];
int n;
int f[maxn][maxn];
unsigned int hash[maxn][maxn];
unsigned short hhhh[maxn][maxn];
void pre_work() {
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      hash[i][j] = hash[i][j - 1] * 100 + s[j] - 'a';
      hash[i][j] %= mod;
      hhhh[i][j] = hhhh[i][j - 1] * 100 + s[j] - 'a';
    }
  }
  for (int i = 0; i < n; i++) {
    int l = i, r = i;
    while (s[l] == s[r] && l >= 0 && r < n) {
      f[l][r] = 1;
      l--;
      r++;
    }
  }
  for (int i = 1; i < n; i++) {
    int l = i - 1, r = i;
    while (s[l] == s[r] && l >= 0 && r < n) {
      f[l][r] = 1;
      l--;
      r++;
    }
  }
}
pair<unsigned int, unsigned short> Hash(int l, int r) {
  return pair<unsigned int, unsigned short>(hash[l][r], hhhh[l][r]);
}
int num[maxn];
void find(int l, int r, int x) {
  if (l > r) return;
  if (f[l][r]) {
    num[f[l][r] + x]++;
  }
  int len = r - l + 1;
  int mid = len >> 1;
  pair<unsigned int, unsigned short> tt1 = Hash(l, l + mid - 1);
  pair<unsigned int, unsigned short> tt2 = Hash(r - mid + 1, r);
  if (tt1.first != tt2.first) return;
  if (tt1.second != tt2.second) return;
  find(l, l + mid - 1, x + 1);
}
void work() {
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      find(i, j, 0);
    }
  for (int i = 1; i <= n; i++) printf("%d ", num[i]);
  puts("");
}
}  // namespace MY
int main() {
  scanf(" %s", MY::s);
  MY::n = strlen(MY::s);
  MY::pre_work();
  MY::work();
  return 0;
}