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;
void read(int &val) {
  int x = 0;
  int bz = 1;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    bz = -1;
    c = getchar();
  }
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  val = x * bz;
}
const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 100;
const int mod = 998244353;
int n, m, a[maxn];
char s[maxn];
unsigned long long hs[maxn], hs2[maxn];
const unsigned long long seed = 5433;
unsigned long long f[maxn];
unsigned long long get(int l, int r) {
  return hs[l] - hs[r + 1] * f[r - l + 1];
}
unsigned long long get2(int l, int r) {
  return hs2[r] - hs2[l - 1] * f[r - l + 1];
}
int dp[5100][5100];
int ans[maxn];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  f[0] = 1;
  for (int i = 1; i <= n; i++) f[i] = f[i - 1] * seed;
  for (int i = n; i >= 1; i--) hs[i] = hs[i + 1] * seed + s[i] - 'a' + 1;
  for (int i = 1; i <= n; i++) hs2[i] = hs2[i - 1] * seed + s[i] - 'a' + 1;
  for (int i = 1; i <= n; i++) dp[i][i] = 1, ans[1]++;
  for (int len = 2; len <= n; len++) {
    for (int l = 1; l <= n - len + 1; l++) {
      int r = l + len - 1;
      int mid = l + (r - l + 1) / 2 - 1, mid2 = r - (mid - l + 1) + 1;
      if (get(l, mid) == get2(mid2, r)) {
        dp[l][r] = dp[l][mid] + 1;
        ans[dp[l][r]]++;
      }
    }
  }
  for (int i = n - 1; i >= 1; i--) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}