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 = 5010;
const long long x = 131;
const long long MOD = 1e9 + 7;
char s[N];
int dp[N][N];
long long H[N];
long long powx[N];
int num[N];
inline long long Hash(int i, int j) {
  return ((H[j] - H[i - 1] * powx[j - i + 1] % MOD) % MOD + MOD) % MOD;
}
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = 1; i <= n; ++i) H[i] = H[i - 1] * x % MOD + s[i], H[i] %= MOD;
  powx[0] = 1;
  for (int i = 1; i <= n; ++i) powx[i] = powx[i - 1] * x % MOD;
  for (int i = 1; i <= n; ++i) dp[i][i] = 1, num[1]++;
  for (int l = 2; l <= n; ++l)
    for (int i = 1; i + l - 1 <= n; ++i) {
      int j = i + l - 1, len = l / 2;
      if (s[i] == s[j] && (i + 1 > j - 1 || dp[i + 1][j - 1])) {
        dp[i][j] = 1;
        if (Hash(i, i + len - 1) == Hash(j - len + 1, j))
          dp[i][j] = dp[i][i + len - 1] + 1;
        num[dp[i][j]]++;
      }
    }
  for (int i = n; i >= 1; --i) num[i] += num[i + 1];
  printf("%d", num[1]);
  for (int i = 2; i <= n; ++i) printf(" %d", num[i]);
}