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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class C {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int inf = Integer.MAX_VALUE;
        String s = in.readLine();
        int n = s.length();
        int[][] next = new int[n][26];
        for (int i = 0; i < n; i++)
            Arrays.fill(next[i], inf);
        for (int i = n - 2; i >= 0; i--) {
            for (int j = 0; j < 26; j++)
                next[i][j] = next[i + 1][j];
            next[i][s.charAt(i + 1) - 'a'] = i + 1;
        }
        long[] res = new long[27];
        int d = 0;
        for (int i = 0; i < n; i++) {
            boolean[] seen = new boolean[26];
            seen[s.charAt(i) - 'a'] = true;
            int cnt = 1;
            int current = i;
            while (true) {
                d = Math.max(d, cnt);
                int nextMin = inf;
                for (int j = 0; j < 26; j++)
                    if (!seen[j])
                        nextMin = Math.min(nextMin, next[i][j]);
                if (nextMin == inf) {
                    res[cnt] += n - current;
                    break;
                } else {
                    seen[s.charAt(nextMin) - 'a'] = true;
                    res[cnt] += (nextMin - current);
                    current = nextMin;
                    cnt += 1;
                }
            }
        }
        System.out.println(d);
        for (int i = 1; i <= d; i++)
            System.out.println(res[i]);
    }
}