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 M = 5e3 + 3;
int n, p, dp[M][M], ans[M], cnt[M], now;
string s;
bool mark[M][M], markp[M][M], palin[M][M];
bool ispal(int l, int r) {
  if (l == r || l > r) return true;
  if (markp[l][r]) return palin[l][r];
  markp[l][r] = true;
  if (s[l] == s[r] && ispal(l + 1, r - 1)) palin[l][r] = true;
  return palin[l][r];
}
int mem(int l, int r) {
  if (mark[l][r]) return dp[l][r];
  mark[l][r] = true;
  if (!ispal(l, r)) return dp[l][r] = 0;
  if (r <= l) return dp[l][r] = 1;
  int mid, mid2;
  if ((r - l + 1) % 2) {
    mid = (l + r) / 2 - 1;
    mid2 = mid + 2;
  } else {
    mid = (l + r) / 2;
    mid2 = mid + 1;
  }
  return dp[l][r] = min(mem(l, mid), mem(mid2, r)) + 1;
}
int main() {
  cin >> s;
  n = s.size();
  s = "0" + s;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) cnt[mem(i, j)]++;
  for (int i = n; i >= 1; i--) {
    now += cnt[i];
    ans[i] = now;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}