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 N = 5000 + 10;
char s[N];
int dp[N][N];
int n;
void spread(int l, int r) {
  while (l >= 0 && r < n && s[l] == s[r]) {
    dp[l][r] = 1;
    l--, r++;
  }
}
void fill_it(int i) {
  spread(i - 1, i + 1);
  spread(i, i + 1);
  dp[i][i] = 1;
}
void solve_even(int l, int r) {
  int l2 = r + 1;
  int r2 = l2 + (r - l);
  if (r2 < n) {
    if (dp[l][r2]) dp[l][r2] = max(dp[l][r2], dp[l][r] + 1);
  }
  r2 = l - 1;
  l2 = r2 - (r - l);
  if (l2 >= 0) {
    if (dp[l2][r]) dp[l2][r] = max(dp[l2][r], dp[l][r] + 1);
  }
}
void solve_odd(int l, int r) {
  int l2 = r + 2;
  int r2 = l2 + (r - l);
  if (r2 < n) {
    if (dp[l][r2]) dp[l][r2] = max(dp[l][r2], dp[l][r] + 1);
  }
  r2 = l - 2;
  l2 = r2 - (r - l);
  if (l2 >= 0) {
    if (dp[l2][r]) dp[l2][r] = max(dp[l2][r], dp[l][r] + 1);
  }
}
int ans[N];
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) fill_it(i);
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      solve_even(i, j);
      solve_odd(i, j);
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      ans[dp[i][j]]++;
    }
  for (int i = n; i >= 0; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], " \n"[i == n]);
}