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 PYTHON3):

Solution:
s = input()
n = len(s)

dp = [[0 for i in range(n - le + 1)] for le in range(n + 1)]
ans = [0 for i in range(n + 1)]

for le in range(1, n + 1):
    for l in range(0, n - le + 1):
        r = l + le
        if s[l] != s[r - 1]:
            continue
        if le == 1:
            dp[1][l] = 1
            ans[1] += 1
        elif le == 2:
            ans[2] += 1
            dp[2][l] = 2
        elif dp[le - 2][l + 1]:
            v = 1
            m = (l + r) // 2
            st = m + 1 if le & 1 else m
            le2 = m - l
            q = dp[le2][l]
            if q:
                v = q + 1

            ans[v] += 1
            dp[le][l] = v


for i in range(n - 1, 0, -1):
    ans[i] += ans[i + 1]

print(*ans[1:])