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;
char sz[5100];
int ans[5100];
int dp[5100][5100], hw[5100][5100];
int len;
int main() {
  scanf("%s", sz + 1);
  len = strlen(sz + 1);
  for (int i = 1; i <= len; i++) hw[i][i] = 1;
  for (int i = 1; i <= len; i++) {
    for (int j = 1; j <= len; j++) {
      if (i - j < 0 || i + j > len) break;
      if (sz[i - j] == sz[i + j]) {
        hw[i - j][i + j] = 1;
      } else
        break;
    }
    for (int j = 1; j <= len; j++) {
      if (i - j + 1 < 0 || i + j > len) break;
      if (sz[i - j + 1] == sz[i + j]) {
        hw[i - j + 1][i + j] = 1;
      } else
        break;
    }
  }
  for (int i = len; i >= 1; i--) {
    for (int j = i; j <= len; j++) {
      if (i == j) {
        ans[1]++;
        dp[i][j] = 1;
      } else {
        int k = (i + j) / 2;
        if ((j - i + 1) % 2 == 1) {
          if (hw[i][j] == 1) {
            dp[i][j] = dp[i][k - 1] + 1;
            for (int p = 1; p <= dp[i][j]; p++) ans[p]++;
          }
        } else {
          if (hw[i][j] == 1) {
            dp[i][j] = dp[i][k] + 1;
            for (int p = 1; p <= dp[i][j]; p++) ans[p]++;
          }
        }
      }
    }
  }
  for (int i = 1; i < len; i++) printf("%d ", ans[i]);
  printf("%d\n", ans[len]);
  return 0;
}