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;
int len, ans[5005];
bool dp[2][5005][5005], par[5005][5005];
char s[5005];
int main() {
  scanf(" %s", s + 1);
  len = strlen(s + 1);
  auto last = dp[0], cur = dp[1];
  for (int i = len; i; --i) {
    cur[i][i] = 1;
    if (i < len) cur[i][i + 1] = (s[i] == s[i + 1]);
    for (int j = i + 2; j <= len; ++j)
      cur[i][j] = (s[i] == s[j]) && cur[i + 1][j - 1];
  }
  memcpy(par, cur, sizeof(par));
  for (int i = 1; i <= len; ++i)
    for (int j = i; j <= len; ++j) ans[1] += cur[i][j];
  for (int k = 2; k <= len; ++k) {
    if (!ans[k - 1]) break;
    swap(cur, last);
    memset(cur, 0, sizeof(dp) >> 1);
    for (int i = 1; i <= len; ++i)
      for (int j = i - 1 + (1 << (k - 1)); j <= len; ++j) {
        if ((j - i) & 1)
          cur[i][j] =
              par[i][j] && last[i][i + j >> 1] && last[(i + j >> 1) + 1][j];
        else
          cur[i][j] = par[i][j] && last[i][(i + j >> 1) - 1] &&
                      last[(i + j >> 1) + 1][j];
        if (cur[i][j]) ++ans[k];
      }
  }
  for (int i = 1; i <= len; ++i) printf("%d%c", ans[i], i == len ? '\n' : ' ');
  return 0;
}