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.Collections;
import java.util.Scanner;


public class C {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String s = in.next();
		long d[] = new long[27];
		ArrayList<Pair> last = new ArrayList();
		for(int i = 1; i <= 26; i++){
			last.add(new Pair(i, -1));
		}
		for(int i = 0; i < s.length(); i++){
			int c = s.charAt(i) - 'a' +1;
			for(int k  = 0; k < last.size(); k++){
				if(last.get(k).c == c){
					last.get(k).last = i;
					break;
				}
			}
			Collections.sort(last);
			for(int j  = 0; j < 26; j++){
				if(j== 0){
					d[26-j] += last.get(j).last + 1;
					continue;
				}
				d[26-j] += last.get(j).last - last.get(j-1).last;
			}
		}
		int max = 0;
		for(int i = 1; i <= 26; i++ ){
			if(d[i] > 0) max = i;
		}
		System.out.println(max);
		for(int i = 1; i <= max; i++){
			System.out.println(d[i]);
		}
	}
	
	static class Pair implements Comparable<Pair>{
		int c;
		int last;
		Pair(int cc , int l){
			c = cc;
			last = l;
		}
		@Override
		public int compareTo(Pair o) {
			// TODO Auto-generated method stub
			return Integer.compare(last, o.last);
		}
	}

}