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 double EPS = 1e-10;
const long long MOD = 1e9 + 7;
const long double PI = 3.1415926535897932384626433832795028841;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
const int maxn = 1e3 * 5 + 20;
const int maxm = 1e5 + 5;
const int maxk = 1e5 + 5;
short int l[maxn][maxn];
bool is_pal[maxn][maxn];
bool new_pal[maxn][maxn];
int ans[maxn];
int main() {
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    l[i][n - 1] = (s[i] == s[n - 1]);
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i - 1; j >= 0; j--) {
      if (s[j] == s[i]) {
        l[j][i] = 1 + l[j + 1][i + 1];
      } else {
        l[j][i] = 0;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    is_pal[i][i] = 1;
    if (i < n - 1) {
      is_pal[i][i + 1] = (s[i] == s[i + 1]);
    }
  }
  for (int sz = 3; sz <= n; sz++) {
    for (int i = 0; i + sz - 1 < n; i++) {
      if (s[i] == s[i + sz - 1]) {
        is_pal[i][i + sz - 1] = is_pal[i + 1][i + sz - 2];
      } else {
        is_pal[i][i + sz - 1] = 0;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      ans[1] += is_pal[i][j];
    }
  }
  for (int kth = 2; kth <= min(n, 15); kth++) {
    for (int i = 0; i < n; i++) {
      for (int j = i; j < n; j++) {
        int sz = j - i + 1;
        int half_sz = sz / 2;
        if (l[i][j - half_sz + 1] >= half_sz) {
          new_pal[i][j] =
              is_pal[i][i + half_sz - 1] && is_pal[j - half_sz + 1][j];
        } else {
          new_pal[i][j] = 0;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = i; j < n; j++) {
        ans[kth] += new_pal[i][j];
        is_pal[i][j] = new_pal[i][j];
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
}