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 inf = numeric_limits<int>::max(), MAX = 5e3 + 5, MOD = 1e9 + 7;
int dp[MAX][MAX];
int ans[MAX];
int lvl[MAX][MAX];
string s;
bool isPalin(int x, int y) {
  if (x >= y) return true;
  if (dp[x][y] != -1) return dp[x][y];
  return (dp[x][y] = (s[x] == s[y] and isPalin(x + 1, y - 1)));
}
int level(int x, int y) {
  if (x == y) return 1;
  if (lvl[x][y] != -1) return lvl[x][y];
  if (!isPalin(x, y)) return lvl[x][y] = 0;
  int mid = (x + y) / 2;
  if ((y - x + 1) % 2 == 0)
    lvl[x][y] = 1 + level(x, mid);
  else
    lvl[x][y] = 1 + level(x, mid - 1);
  return lvl[x][y];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  memset(dp, -1, sizeof dp);
  memset(lvl, -1, sizeof lvl);
  cin >> s;
  int cnt = 0;
  for (int i = 0; i < s.length(); ++i)
    for (int j = i; j < s.length(); ++j) {
      cnt = level(i, j);
      ans[1]++;
      ans[cnt + 1]--;
    }
  for (int i = 1; i <= s.length(); ++i) {
    ans[i] += ans[i - 1];
    cout << ans[i] << " ";
  }
  cout << "\n";
  return 0;
}