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 INF = 0x3f3f3f3f;
const int N = 5e3 + 10;
char s[N];
bool f[N][N];
int dp[N][N];
int cnt[N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  memset(f, false, sizeof(f));
  for (int i = 1; i <= n; i++) f[i][i] = f[i][i - 1] = true;
  for (int l = 2; l <= n; l++) {
    for (int i = 1; i <= n - l + 1; i++) {
      int j = i + l - 1;
      f[i][j] = (f[i + 1][j - 1] && s[i] == s[j]);
    }
  }
  for (int i = 1; i <= n; i++) dp[i][i] = dp[i][i - 1] = true;
  for (int l = 2; l <= n; l++) {
    for (int i = 1; i <= n - l + 1; i++) {
      int j = i + l - 1;
      if (!f[i][j]) continue;
      int lm = i + l / 2 - 1;
      int rm = j - l / 2 + 1;
      dp[i][j] = min(dp[i][lm], dp[rm][j]) + 1;
    }
  }
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) cnt[dp[i][j]]++;
  for (int i = n - 1; i >= 1; i--) cnt[i] += cnt[i + 1];
  for (int i = 1; i < n; i++) printf("%d ", cnt[i]);
  printf("%d\n", cnt[n]);
  return 0;
}