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 = 5010;
const double eps = 1e-8;
const int iinf = INT_MAX;
const long long linf = 2e18;
const double dinf = 1e30;
const int MOD = 1000000007;
inline int read() {
  int X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
void print(int x) {
  cout << x << '\n';
  exit(0);
}
void PRINT(string x) {
  cout << x << '\n';
  exit(0);
}
void douout(double x) { printf("%lf\n", x + 0.0000000001); }
char ch[N];
int ans[N], dp[N][N];
int main() {
  while (~scanf("%s", ch + 1)) {
    int len = strlen(ch + 1);
    memset(ans, 0, sizeof ans);
    memset(dp, 0, sizeof dp);
    for (int i = 1; i <= len; i++) {
      dp[i][i] = 1;
      ans[1]++;
      if (i < len && ch[i] == ch[i + 1]) dp[i][i + 1] = 2, ans[2]++, ans[1]++;
    }
    for (int i = 3; i <= len; i++)
      for (int j = 1; j + i - 1 <= len; j++)
        if (dp[j + 1][j + i - 2] && ch[j] == ch[i + j - 1]) {
          dp[j][j + i - 1] = dp[j][j + i / 2 - 1] + 1;
          for (int k = 1; k <= dp[j][j + i - 1]; k++) ans[k]++;
        }
    printf("%d", ans[1]);
    for (int i = 2; i <= len; i++) printf(" %d", ans[i]);
    printf("\n");
  }
  return 0;
}