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 = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
string s;
bool memo[5000][5000];
bool dp[5000][5000][2];
bool used[5000][5000][2];
int hash_val[5000][5000];
int cnt[5000];
bool dfs(int a, int b) {
  if (a == b || b < a) return dp[a][b][1] = 1;
  if (used[a][b][1]) return dp[a][b][1];
  used[a][b][1] = 1;
  dp[a][b][1] = 0;
  if (s[a] == s[b]) dp[a][b][1] = dfs(a + 1, b - 1);
  return dp[a][b][1];
}
bool dfs(int a, int b, int k) {
  if (used[a][b][k]) return dp[a][b][k];
  used[a][b][k] = 1;
  int len = b - a + 1;
  int en, st;
  en = a + len / 2 - 1;
  if (!dp[a][en][k - 1]) return dp[a][b][k] = 0;
  st = b - len / 2 + 1;
  if (!dp[st][b][k - 1]) return dp[a][b][k] = 0;
  if (!memo[a][b]) return dp[a][b][k] = 0;
  return dp[a][b][k] = 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  while (cin >> s) {
    n = s.length();
    memset(memo, 0, sizeof(memo));
    memset(dp, 0, sizeof(dp));
    memset(used, 0, sizeof(used));
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n; ++i) {
      for (int j = i; j < n; ++j) {
        if (dfs(i, j)) {
          cnt[1]++;
          memo[i][j] = 1;
        }
      }
    }
    for (int k = 2; (1 << (k - 1)) <= n; ++k) {
      for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
          dp[i][j][0] = dp[i][j][1];
        }
      }
      memset(used, 0, sizeof(used));
      for (int i = 0; i < n; ++i) {
        for (int j = i + (1 << (k - 1)) - 1; j < n; ++j) {
          if (dfs(i, j, 1)) cnt[k]++;
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      cout << cnt[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}