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 + 50;
char s[maxn];
long long pre[5000 + 50];
long long H[5005];
int num[maxn][maxn];
int main() {
  memset(num, 0, sizeof(num));
  memset(H, 0, sizeof(H));
  scanf("%s", s);
  int d = strlen(s);
  for (int i = 0; i < d; i++) {
    for (int p = 0; i - p >= 0 && i + p < d; p++) {
      if (s[i - p] == s[i + p]) {
        num[i - p][i + p] = 1;
      } else
        break;
    }
    for (int p = 0; i - p >= 0 && i + 1 + p < d; p++) {
      if (s[i - p] == s[i + 1 + p])
        num[i - p][i + 1 + p] = 1;
      else
        break;
    }
  }
  for (int i = 1; i <= d; i++) {
    for (int h = 0; h < d; h++) {
      if (i == 1) {
        num[h][h + i - 1] = 1;
        H[num[h][h + i - 1]]++;
        continue;
      }
      if (h + i - 1 < d) {
        if (num[h][h + i - 1]) {
          num[h][h + i - 1] = num[h][h + i / 2 - 1] + 1;
          H[num[h][h + i - 1]]++;
        }
      }
    }
  }
  stack<int> G;
  long long ans = 0;
  for (int p = d; p >= 1; p--) {
    G.push(H[p] + ans);
    ans += H[p];
  }
  while (!G.empty()) {
    int k = G.top();
    cout << k << ' ';
    G.pop();
  }
  cout << endl;
  return 0;
}