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 MAXN = 5010;
const long long BASE = 61;
const long long MOD = 1e9 + 7;
long long dp[MAXN][MAXN], pw[MAXN], hs[MAXN];
bool visit[MAXN][MAXN];
long long ans[MAXN];
int get_hash(int l, int r) {
  return (((hs[r] - hs[l - 1] * pw[r - l + 1]) % MOD) + MOD) % MOD;
}
long long find(int l, int r) {
  if (visit[l][r]) return dp[l][r];
  visit[l][r] = true;
  int len = (r - l + 1) / 2;
  long long h1 = get_hash(l, l + len - 1);
  long long h2 = get_hash(r - len + 1, r);
  if (h1 == h2) {
    find(l, l + len - 1);
    return dp[l][r] += find(r - len + 1, r);
  } else {
    return dp[l][r] = 1;
  }
}
int main(void) {
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  s = " " + s;
  pw[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    pw[i] = (pw[i - 1] * BASE) % MOD;
  }
  for (int i = 1; i < s.length(); i++) {
    hs[i] = (hs[i - 1] * BASE + s[i] - 'a' + 1) % MOD;
  }
  for (int i = 1; i < s.length(); i++) {
    dp[i][i] = 1;
    visit[i][i] = true;
    if (i < s.length() - 1 && s[i] == s[i + 1]) {
      dp[i][i + 1] = 2;
      visit[i][i + 1] = true;
    }
  }
  for (int len = 2; len < s.length(); len++) {
    for (int i = 1; i + len - 1 < s.length(); i++) {
      if (s[i] == s[i + len - 1] && dp[i + 1][i + len - 2]) {
        dp[i][i + len - 1] = 1;
      }
    }
  }
  for (int len = s.length() - 1; len > 2; len--) {
    for (int i = 1; i + len - 1 < s.length(); i++) {
      if (dp[i][i + len - 1] > 0 && !visit[i][i + len - 1]) {
        find(i, i + len - 1);
      }
    }
  }
  for (int len = s.length() - 1; len > 0; len--) {
    for (int i = 1; i + len - 1 < s.length(); i++) {
      if (dp[i][i + len - 1]) {
        ans[dp[i][i + len - 1]]++;
      }
    }
  }
  for (int i = s.length() - 1; i > 0; i--) {
    ans[i] += ans[i + 1];
  }
  for (int i = 1; i < s.length(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}