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>
const long long N = 5050;
const long long Q = 101;
const long long mod = 1e9 + 7;
const long long block = 250;
using namespace std;
int n;
int k[N];
int a[N][N];
int d[N][N];
bool p[N][N];
string s;
bool P(int l, int r) { return p[l][r]; }
void solve() {
  cin >> s;
  n = s.size();
  string t = s;
  reverse(t.begin(), t.end());
  for (int i = n - 1; i >= 0; i--) {
    for (int j = n - 1; j >= 0; j--) {
      if (s[i] == s[j]) {
        a[i][j] = a[i + 1][j + 1] + 1;
      }
      if (i == j) {
        p[i][j] = 1;
      } else if (s[i] == s[j] && (i + 1 == j || p[i + 1][j - 1])) {
        p[i][j] = 1;
      }
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i; j < n; j++) {
      k[1] += P(i, j);
      d[i][j] = P(i, j);
      if (i == j) {
        continue;
      }
      int g = (j - i + 1) / 2;
      int mn = min(d[i][i + g - 1], d[j - g + 1][j]);
      if (a[i][j - g + 1] >= g) {
        d[i][j] = mn + 1;
      }
      if (d[i][j] <= 1) {
        d[i][j] = P(i, j);
      } else {
        k[d[i][j]]++;
      }
    }
  }
  for (int i = n; i >= 2; i--) {
    k[i] += k[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    cout << k[i] << " ";
  }
}
bool mtest = false;
int main() {
  ios_base::sync_with_stdio(0);
  int TE = 1;
  if (mtest) cin >> TE;
  while (TE--) solve();
  return 0;
}