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.awt.*;
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Abc {
    static String s;
    static int dp[][];
    static int n;
    static int cnt[];
    public static void main(String[] args) throws Exception {
        Scanner sc=new Scanner(System.in);
        s=sc.next();
        n=s.length();
        dp=new int[n][n];
        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++){
                dp[i][j]=-1;
            }
        }
        cnt=new int[n+1];
        for (int i=0;i<n;i++){
            for (int j=i;j<n;j++){
                int x=dp(i,j);
                cnt[x]++;
            }
        }
        for (int i=n-1;i>=1;i--){
            cnt[i]+=cnt[i+1];
        }
        StringBuilder sb=new StringBuilder();
        for (int i=1;i<=n;i++){
            sb.append(cnt[i]+" ");
        }
        System.out.println(sb);
    }
    static int dp(int i,int j){
        if (j<=i)return 1;
        if (s.charAt(i)!=s.charAt(j))return 0;
        if (dp[i][j]!=-1)return dp[i][j];
        int len=(j-i+1)/2;
        if(dp(i+1,j-1)>0){
            dp[i][j]=Math.min(dp(i,i+len-1),dp(j-len+1,j))+1;
        }else return dp[i][j]=0;
        return dp[i][j];
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}