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 MOD = 998244353, SHIFT = 29;
int n, m;
char buf[5010];
int Hash[5010][5010];
short palin[5010][5010];
int ans[15];
int main() {
  scanf("%s", buf);
  int len = strlen(buf);
  for (int i = 0; i < len; i++) {
    for (int j = i; j < len; j++) {
      Hash[i][j] = ((long long)Hash[i][j - 1] * SHIFT + buf[j] - 'a' + 1) % MOD;
    }
  }
  for (int i = 0; i < len; i++) {
    for (int j = i - 1; j >= 0; j--) {
      Hash[i][j] = ((long long)Hash[i][j + 1] * SHIFT + buf[j] - 'a' + 1) % MOD;
    }
  }
  for (int i = 0; i < len; i++) {
    for (int j = i; j < len; j++) {
      if (Hash[i][j] == Hash[j][i]) {
        palin[i][j] = 1;
        ans[1]++;
      }
    }
  }
  for (int l = 2; l <= 13; l++) {
    for (int i = 0; i < len; i++) {
      for (int j = i + 1; j < len; j++) {
        int d = (j - i - 1) / 2;
        if (palin[i][j] >= l - 1 && palin[i][i + d] >= l - 1 &&
            palin[j - d][j] >= l - 1) {
          ans[l]++;
          palin[i][j] = l;
        }
      }
    }
  }
  for (int i = 1; i <= len; i++) {
    if (i <= 13)
      printf("%d ", ans[i]);
    else
      printf("0 ");
  }
  return 0;
}