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;

/**
 * Runtime O(12*n^2), Space O(n^2)
 *
 * @author Don Li
 */
public class PalindromicCharacteristics {
    
    void solve() {
        char[] s = in.nextToken().toCharArray();
        int n = s.length;
        
        int[][] lcp = new int[n + 1][n + 1];
        for (int i = n - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                lcp[i][j] = lcp[i + 1][j + 1] + (s[i] == s[j] ? 1 : 0);
            }
        }
        
        int[][] dp = new int[n][n];
        
        for (int i = 0; i < n; i++) {
            for (int x = 0; x <= Math.min(i, n - i - 1); x++) {
                if (s[i - x] == s[i + x]) dp[i - x][i + x] = 1;
                else break;
            }
        }
        for (int i = 0; i < n - 1; i++) {
            for (int x = 0; x <= Math.min(i, n - i - 2); x++) {
                if (s[i - x] == s[i + 1 + x]) dp[i - x][i + 1 + x] = 1;
                else break;
            }
        }
    
        for (int k = 1; k <= 12; k++) {
            for (int l = 0; l < n; l++) {
                for (int r = l; r < n; r++) {
                    if (dp[l][r] >= k) {
                        int m = (l + r) >> 1;
                        if (((r - l + 1) & 1) == 0) {
                            if (l <= m && dp[l][m] >= dp[l][r] && m + 1 <= r && dp[m + 1][r] >= dp[l][r] && lcp[l][m + 1] >= (r - l + 1) >> 1) {
                                dp[l][r]++;
                            }
                        } else {
                            if (l <= m - 1 && dp[l][m - 1] >= dp[l][r] && m + 1 <= r && dp[m + 1][r] >= dp[l][r] && lcp[l][m + 1] >= (r - l) >> 1) {
                                dp[l][r]++;
                            }
                        }
                    }
                }
            }
        }
        
        // calc the result using partial_sum
        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 PalindromicCharacteristics().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());
        }
    }
}