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>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
inline void read(int &x) {
  x = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + (c - '0');
}
inline void write(int x) {
  register char buffor[35];
  register int i = 0;
  do {
    buffor[i++] = (x % 10) + '0';
    x /= 10;
  } while (x);
  i--;
  while (i >= 0) putchar(buffor[i--]);
  putchar('\n');
}
inline int in() {
  int x;
  read(x);
  return x;
}
void solve();
int dbg = 1;
using namespace std;
int main() {
  int t = 1;
  for (int i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
int dp[5005][5005];
int dp3[5005][5005];
string second;
void chec(int l, int r) {
  if (l == r) {
    dp[l][r] = 1;
    return;
  } else if (dp[l][r] != -1)
    return;
  int rst = (r - l + 2) / 2, len = r - l + 1;
  chec(l, l + len / 2 - 1);
  chec(rst + l, r);
  int f2 = dp3[l][r];
  if (f2) dp[l][r] = 1;
  if (dp[l][l + len / 2 - 1] != -1 && dp[l + rst][r] != -1 && f2)
    dp[l][r] = dp[l + rst][r] + 1;
  dp[l][r] = max(dp[l][r], 0);
}
int func(int l, int r) {
  if (l == r) {
    dp3[l][r] = 1;
    return 1;
  } else if (dp3[l][r] != -1)
    return dp3[l][r];
  if (second[l] == second[r])
    if (l + 1 == r)
      return dp3[l][r] = 1;
    else
      return dp3[l][r] = func(l + 1, r - 1);
  return dp3[l][r] = 0;
}
int arr[5005];
void solve() {
  cin >> second;
  memset(dp, -1, sizeof(dp));
  memset(dp3, -1, sizeof(dp3));
  int n = second.length();
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      func(i, j);
    }
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      chec(i, j);
      arr[dp[i][j]]++;
    }
  for (int i = n - 1; i >= 0; i--) arr[i] += arr[i + 1];
  for (int i = 1; i <= n; i++) cout << arr[i] << " ";
  cout << endl;
}