Problem: String diversity is the number of symbols that occur in the string at least once. Diversity of s will be denoted by d(s). For example , d("aaa")=1, d("abacaba")=3.

Given a string s, consisting of lowercase Latin letters. Consider all its substrings. Obviously, any substring diversity is a number from 1 to d(s). Find statistics about substrings diversity: for each k from 1 to d(s), find how many substrings of s has a diversity of exactly k.

Input

The input consists of a single line containing s. It contains only lowercase Latin letters, the length of s is from 1 to 3·105.

Output

Print to the first line the value d(s). Print sequence t1, t2, ..., td(s) to the following lines, where ti is the number of substrings of s having diversity of exactly i.

Examples

Input

abca


Output

3
4
3
3


Input

aabacaabbad


Output

4
14
19
28
5

Note

Consider the first example.

We denote by s(i, j) a substring of "abca" with the indices in the segment [i, j].

  * s(1, 1) =  "a", d("a") = 1
  * s(2, 2) =  "b", d("b") = 1
  * s(3, 3) =  "c", d("c") = 1
  * s(4, 4) =  "a", d("a") = 1
  * s(1, 2) =  "ab", d("ab") = 2
  * s(2, 3) =  "bc", d("bc") = 2
  * s(3, 4) =  "ca", d("ca") = 2
  * s(1, 3) =  "abc", d("abc") = 3
  * s(2, 4) =  "bca", d("bca") = 3
  * s(1, 4) =  "abca", d("abca") = 3



Total number of substring with diversity 1 is 4, with diversity 2 equals 3, 3 diversity is 3.

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static final int N = 27;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        Main main = new Main();
        for(Long i: main.count(sc.next().toCharArray())){
            System.out.println(i);
        }
    }

    List<Long> count(char[] input){
        int k = getK(input);
        long[][] lengths = new long[k][input.length];
        int[][] masks = new int[k][input.length];
        lengths[0][0] = 1;
        masks[0][0] = mask(input[0]);
        for (int i = 1; i < input.length; i++) {
            for (int j = k - 1; j >= 0; j--) {
                int m = mask(input[i]);
                if(j == 0){
                    lengths[j][i] = 1;
                    masks[j][i]= m;
                } else if((masks[j - 1][i - 1] & m) == 0) {
                    lengths[j][i] += lengths[j - 1][i - 1];
                    masks[j][i] = masks[j - 1][i - 1] | m;
                }
                if((masks[j][i - 1] & m) > 0){
                    lengths[j][i] += lengths[j][i - 1];
                    masks[j][i] = masks[j][i - 1] | m;
                }
            }
        }
        List<Long> result = new ArrayList<>();
        result.add((long)k);
        for (int i = 0; i < k; i++) {
            long row = 0;
            for (int j = 0; j < input.length; j++) {
                row += lengths[i][j];
            }
            result.add(row);
        }
        return result;
    }

    private int mask(char c){
        return 1 << (c - 'a');
    }

    private int getK(char[] input) {
        int[] set = new int[N];
        for (int i = 0; i < input.length; i++) {
            set[input[i] - 'a']++;
        }
        int count = 0;
        for (int i = 0; i < set.length; i++) {
            count += set[i] == 0 ? 0 : 1;
        }
        return count;
    }
}