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;
long long dp[5005][5005], cnt[5005];
char s[5005];
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
signed main() {
  cin >> (s + 1);
  long long n = strlen(s + 1);
  for (long long i = 1; i <= n; i++) {
    dp[i][i] = 1;
    if (s[i] == s[i + 1]) dp[i][i + 1] = 2, cnt[2]++, cnt[1]++;
    cnt[1]++;
  }
  for (long long i = 3; i <= n; i++) {
    for (long long l = 1; l + i - 1 <= n; l++) {
      long long r = l + i - 1;
      if (!dp[l + 1][r - 1] || s[l] != s[r]) continue;
      dp[l][r] = dp[l][l + i / 2 - 1] + 1;
      for (long long j = 1; j <= dp[l][r]; j++) cnt[j]++;
    }
  }
  for (long long i = 1; i <= n; i++) cout << cnt[i] << ' ';
  return 0;
}