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 long long int N = 5e3 + 5, mod = 1e9 + 7, bit = 60, p = 199;
long long int pw[N], hash1[N], hash2[N], n, ans[N] = {0}, inv[N] = {0},
                                            dp[N][N] = {0};
string s;
long long int power(long long int a, long long int b, long long int m = mod) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % m;
    }
    b >>= 1;
    a = (a * a) % m;
  }
  return res;
}
void pre() {
  pw[0] = 1;
  for (int i = 1; i < N; i++) {
    pw[i] = (pw[i - 1] * p) % mod;
  }
  inv[N - 1] = power(pw[N - 1], mod - 2);
  for (int i = N - 2; i >= 0; i--) {
    inv[i] = (inv[i + 1] * p) % mod;
  }
}
void make() {
  long long int i;
  hash1[0] = 0;
  hash2[n + 1] = 0;
  for (i = 1; i <= n; i++) {
    hash1[i] = (hash1[i - 1] + (s[i] - 'a') * (pw[i])) % mod;
  }
  for (i = n; i >= 1; i--) {
    hash2[i] = (hash2[i + 1] + (s[i] - 'a') * (pw[n - i + 1])) % mod;
  }
}
long long int get1(long long int l, long long int r) {
  if (l > r) {
    return 0;
  }
  long long int ans = (hash1[r] - hash1[l - 1] + mod) % mod;
  ans = (ans * inv[l - 1]) % mod;
  return ans;
}
long long int get2(long long int l, long long int r) {
  if (l > r) {
    return 0;
  }
  long long int ans = (hash2[l] - hash2[r + 1] + mod) % mod;
  ans = (ans * inv[n - r]) % mod;
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int i, j, l, r, k;
  cin >> s;
  n = s.length();
  s = '@' + s;
  pre();
  make();
  for (i = 1; i <= n; i++) {
    dp[i][i] = 1;
    ans[dp[i][i]]++;
  }
  for (i = 1; i < n; i++) {
    if (s[i] == s[i + 1]) {
      dp[i][i + 1] = 2;
    }
    ans[dp[i][i + 1]]++;
  }
  for (i = 3; i <= n; i++) {
    for (j = 1; j + i - 1 <= n; j++) {
      long long int mid = (i) >> 1;
      mid--;
      if (get1(j, j + i - 1) == get2(j, j + i - 1)) {
        dp[j][j + i - 1] = 1;
      }
      if (get1(j, j + mid) == get1(j + i - 1 - mid, j + i - 1) and
          get1(j, j + mid) == get2(j, j + mid)) {
        dp[j][j + i - 1] = 1 + dp[j][j + mid];
      }
      ans[dp[j][j + i - 1]]++;
    }
  }
  for (i = n - 1; i >= 1; i--) {
    ans[i] += ans[i + 1];
  }
  for (i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
  return 0;
}