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;
long long int mod = 998244353;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int m = 0, n, res, k, ans = 0, nbcycle = 0;
string s2 = "zero", ch = "", s, t, s1 = "";
set<pair<pair<int, int>, int> > st;
vector<pair<string, int> > vm;
vector<pair<int, pair<int, int> > > vvv;
vector<pair<int, int> > vv, vv2;
vector<int> v5;
map<int, int> mp;
bool dp[2][5005][5005], b[5005][5005], pal[5005][5005];
int a[5005] = {0};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int i = 0, x, y, j = 0, q, mm = 0, mx = 0, idx = 0, ok = 0, l = 0, r = 0,
      negatif = 0, positif = 0, l1, r1, d;
  set<pair<pair<int, int>, int> >::iterator it, it1;
  cin >> s;
  string s1 = "";
  n = s.size();
  for (j = 0; j < n; j++) {
    pal[i][i] = 1;
    for (i = j; i >= 0; i--) {
      if (i == j)
        dp[0][i][j] = 1;
      else if (i == (j - 1) && s[i] == s[j])
        dp[0][i][j] = 1;
      else if (s[i] == s[j] && dp[0][i + 1][j - 1])
        dp[0][i][j] = 1;
      a[1] += dp[0][i][j];
      pal[i][j] = dp[0][i][j];
    }
  }
  int flag = 1;
  for (k = 2; k <= min(13, n); k++) {
    flag ^= 1;
    for (j = 0; j < n; j++) {
      for (i = 0; i <= j; i++) {
        if ((j - i + 1) & 1) {
          b[i][j] = 0;
          int mid = (i + j) / 2;
          if (dp[flag][i][mid - 1] && dp[flag][mid + 1][j] && pal[i][j])
            b[i][j] = 1;
          dp[flag ^ 1][i][j] = b[i][j];
          a[k] += b[i][j];
          b[i][j] = 0;
        } else {
          b[i][j] = 0;
          int mid = (i + j) / 2;
          if (dp[flag][i][mid] && dp[flag][mid + 1][j] && pal[i][j])
            b[i][j] = 1;
          dp[flag ^ 1][i][j] = b[i][j];
          a[k] += b[i][j];
          b[i][j] = 0;
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (i > 13) a[i] = 0;
    cout << a[i] << " ";
  }
  return 0;
}