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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class PalindromicCharacteristics2 {
    void solve() {
        char[] s = in.nextToken().toCharArray();
        int n = s.length;
        
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; i++) dp[i][i] = 1;
        for (int i = 0; i < n - 1; i++) if (s[i] == s[i + 1]) dp[i][i + 1] = 2;
        for (int x = 1; x < n; x++) {
            for (int i = 0; i < n; i++) {
                if (i - x < 0 || i + x >= n || s[i - x] != s[i + x] || dp[i - x + 1][i + x - 1] == 0) continue;
                dp[i - x][i + x] = dp[i - x][i - 1] + 1;
            }
            for (int i = 0; i < n - 1; i++) {
                if (i - x < 0 || i + 1 + x >= n || s[i - x] != s[i + 1 + x] || dp[i - x + 1][i + x] == 0) continue;
                dp[i - x][i + 1 + x] = dp[i - x][i] + 1;
            }
        }
        
        int[] res = new int[n + 1];
        for (int i = 0; i < n; i++) for (int j = i; j < n; j++) res[dp[i][j]]++;
        for (int i = n - 1; i > 0; i--) res[i] += res[i + 1];
        
        for (int i = 1; i <= n; i++) {
            if (i > 1) out.print(' ');
            out.print(res[i]);
        }
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new PalindromicCharacteristics2().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}