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 + 10;
int dp[maxn][maxn];
char s[maxn];
long long ans[maxn];
int main() {
  while (~scanf("%s", s)) {
    int len = strlen(s);
    memset(ans, 0, sizeof(ans));
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < len; ++i) {
      dp[i][i] = 1;
      ++ans[1];
    }
    for (int l = 2; l <= len; ++l) {
      for (int i = 0; i + l - 1 < len; ++i) {
        if (((i + 1 > i + l - 2) || dp[i + 1][i + l - 2]) &&
            s[i] == s[i + l - 1])
          dp[i][i + l - 1] = dp[i][i + l / 2 - 1] + 1;
        for (int k = 1; k <= dp[i][i + l - 1]; ++k) ++ans[k];
      }
    }
    for (int i = 1; i <= len; ++i)
      if (i == len)
        printf("%d\n", ans[i]);
      else
        printf("%d ", ans[i]);
  }
  return 0;
}