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.*;

public class C
{
    public static void main(String[] args) throws IOException
    {
        final byte[] buf = new byte[3*100000 + 10];
        int n = System.in.read(buf);
        while(buf[n-1]<=' ') --n;
        
        final int inf = Integer.MAX_VALUE;
        final int[][] next = new int[n][26];
        for(int i = 0; i < n; i++)
            java.util.Arrays.fill(next[i], inf);
        for(int i = n - 2; i >= 0; i--)
        {
            final int[] v = next[i], u = next[i+1];
            for(int j = 0; j < 26; j++)
                v[j] = u[j];
            v[buf[i + 1] - 'a'] = i + 1;
        }
        
        final long[] res = new long[27];
        int d = 0;
        for(int i = 0; i < n; i++)
        {
            final int[] v = next[i];
            
            final boolean[] seen = new boolean[26];
            seen[buf[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, v[j]);
                if(nextMin == inf)
                {
                    res[cnt] += n - current;
                    break;
                }
                else
                {
                    seen[buf[nextMin] - 'a'] = true;
                    res[cnt] += (nextMin - current);
                    current = nextMin;
                    cnt += 1;
                }
            }
        }
        
        final PrintWriter out = new PrintWriter(System.out);
        out.println(d);
        for (int i = 1; i <= d; i++)
            out.println(res[i]);
        out.flush();
    }
}