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.io.*;
import java.util.*;

public class Main {

    public static void main(String args[]) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        while(in.hasNext()) {
            String s = in.next();
            Queue<Integer>[] next = new LinkedList[30];
            long ans[] = new long[30];
            for(int i=0;i<26;i++) {
                next[i] = new LinkedList<Integer>();
            }
            for(int i=0;i<s.length();i++) {
                char c = s.charAt(i);
                next[c-'a'].add(i);
            }
            for(int i=0;i<s.length();i++) {
                int pos[] = new int[30],sz=0;
                for(int j=0;j<26;j++) {
                    if(next[j].size()>0) {
                        pos[sz++]=next[j].peek();
                    }
                }
                if(sz>0) {
                    Arrays.sort(pos,0,sz);
                    int now=pos[0];
                    for(int j=1;j<sz;j++) {
                        ans[j]+=pos[j]-pos[j-1];
                        now=pos[j];
                    }
                    ans[sz]+=s.length()-now;
                }
                for(int j=0;j<26;j++) {
                    if(next[j].size()>0) {
                        int val = next[j].peek();
                        if(val == i) {
                            next[j].poll();
                            break;
                        }
                    }
                }
            }
            int tot=0,use[] = new int[30];
            for(int i=0;i<s.length();i++) {
                char k = s.charAt(i);
                if(use[k-'a']==0) {
                    use[k-'a']=1;
                    tot++;
                }
            }
            System.out.println(tot);
            for(int i=1;i<=tot;i++){
                System.out.println(ans[i]);
            }
        }
    }
}