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 = 5000 + 5;
char s[MAXN + 5];
bool check[MAXN + 5][MAXN + 5];
int f[MAXN + 5][MAXN + 5];
int sumv[MAXN + 5];
int N;
int main() {
  scanf("%s", s + 1);
  N = strlen(s + 1);
  for (int i = 1; i <= N; i++) check[i][i] = true;
  for (int i = 1; i <= N; i++) {
    for (int j = i - 1; j >= 1; j--) {
      if (j + 1 > i - 1 && s[i] == s[j])
        check[j][i] = true;
      else if (check[j + 1][i - 1] && s[j] == s[i])
        check[j][i] = true;
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = i; j <= N; j++)
      if (check[i][j]) f[i][j] = 1;
  }
  for (int L = 2; L <= N; L++) {
    for (int j = 1; j + L - 1 <= N; j++) {
      int a = j + L / 2 - 1;
      int b = j + L - 1 - L / 2 + 1;
      if (check[j][j + L - 1]) {
        int c = f[j][a], d = f[b][j + L - 1];
        if (c > 0 && d > 0) f[j][j + L - 1] = min(c, d) + 1;
      }
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = i; j <= N; j++) {
      int c = f[i][j];
      sumv[c]++;
    }
  }
  for (int i = N; i >= 1; i--) sumv[i] += sumv[i + 1];
  for (int i = 1; i <= N; i++) {
    if (i == N)
      printf("%d\n", sumv[i]);
    else
      printf("%d ", sumv[i]);
  }
}