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 = 5005;
const long long int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long int base = 31337;
long long int pot[N], roll[N][3], sol[N], n, dp[N][N];
string s;
inline long long int aaa(long long int x) {
  if (x < 0) return x + MOD;
  return x;
}
void pre() {
  pot[0] = 1;
  for (int i = 1; i <= n; i++) {
    pot[i] = pot[i - 1] * base % MOD;
  }
  for (int i = 0; i < n; i++) {
    roll[i + 1][0] = (roll[i][0] + s[i] * pot[i] % MOD) % MOD;
  }
  for (int i = n - 1; i >= 0; i--) {
    roll[i + 1][1] = (roll[i + 2][1] + s[i] * pot[n - 1 - i] % MOD) % MOD;
  }
}
long long int gethsh(int i, int j, int sm) {
  if (!sm) {
    return aaa(roll[j][sm] - roll[i - 1][sm]) * pot[n - j + 1] % MOD;
  } else {
    return aaa(roll[i][sm] - roll[j + 1][sm]) * pot[i] % MOD;
  }
}
long long int checkpalin(int i, int j) {
  return gethsh(i, j, 0) == gethsh(i, j, 1);
}
void prijelaz(int i, int j) {
  long long int raz;
  if ((j - i) % 2 &&
      gethsh(i, (i + j) / 2, 0) == gethsh((i + j) / 2 + 1, j, 0)) {
    raz = dp[i][(i + j) / 2] + 1;
  } else if (gethsh(i, (i + j) / 2 - 1, 0) == gethsh((i + j) / 2 + 1, j, 0)) {
    raz = dp[i][(i + j) / 2 - 1] + 1;
  } else {
    raz = 1;
  }
  dp[i][j] = raz;
  sol[dp[i][j] - 1]++;
}
int main() {
  cin >> s;
  n = s.size();
  pre();
  for (int i = 1; i <= n; i++) {
    for (int j = 0; i + j <= n; j++) {
      if (!checkpalin(i, i + j)) continue;
      prijelaz(i, i + j);
    }
  }
  int c = 0;
  for (int i = n; i >= 0; i--) {
    c += sol[i];
    sol[i] = c;
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", sol[i]);
  }
  printf("\n");
}