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 N = 5e3 + 20;
char s[N];
int dp[N][N], ans[N];
int main() {
  while (scanf("%s", s + 1) != EOF) {
    int n = strlen(s + 1);
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) dp[i][i] = 1;
    for (int len = 2; len <= n; len++) {
      for (int i = 1; i + len - 1 <= n; i++) {
        int l = i, r = i + len - 1;
        if (len == 2) {
          if (s[l] == s[r]) dp[l][r] = 2;
          continue;
        }
        int m = len >> 1;
        if (s[l] == s[r] && dp[l + 1][r - 1]) {
          dp[l][r] = 1;
          if (dp[l][l + m - 1] && dp[r - m + 1][r])
            dp[l][r] = dp[l][l + m - 1] + 1;
        }
      }
    }
    memset(ans, 0, sizeof(ans));
    for (int len = 1; len <= n; len++) {
      for (int i = 1; i + len - 1 <= n; i++) {
        ans[dp[i][i + len - 1]]++;
      }
    }
    for (int i = n - 1; i >= 1; i--) ans[i] += ans[i + 1];
    for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}