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.*;
import java.io.*;

public class Solution
{
       public static void main(String []args) throws IOException
       {
             BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
             String s = br.readLine();
             
              int n = s.length();
             int next[][] = new int[26][n];
             for(int i = 0 ; i < 26 ; i++)
             {
                  for(int j = 0 ; j < n ; j++)
                  {
                         next[i][j] = n;
                  }
             }
             
             for(int i = 0 ; i < 26 ; i++)
             {
                    for(int j = n-1 ; j >= 0 ; j--)
                    {
                           if(s.charAt(j) -'a' == i)
                           {
                                  next[i][j] = j;
                           }
                           
                           else
                           {
                                  if(j != n-1)
                                  {
                                         next[i][j] = next[i][j+1];
                                  }
                           }
                    }
             }
              long ans[] = new long[26];
              int ass = 0;
              StringBuffer str = new StringBuffer("");
              for(int k = 1 ; k <= 26 ; k++)
              {
                   int freq[] = new int[26];
                   int tot = 0;
                   int i = 0;
                   int j = 0;
                   while(i < n && j < n)
                   {
                          int ch = s.charAt(j)-'a';
                          if(freq[ch] == 0)
                          {
                                 tot++;
                          }
                          freq[ch]++;
                          if(tot < k)
                          j++;
                          
                          else
                          {
                                 int mx = n;
                            for(int kk = 0 ; kk < 26 ; kk++)
                            {
                                   if(freq[kk]==0)
                                   {
                                          mx = Math.min(mx,next[kk][j]);
                                   }
                            }
                            
                                 while(tot == k)
                                 {
                                 
                            ans[k-1] += mx-j;
                            freq[s.charAt(i)-'a']--;
                            if(freq[s.charAt(i)-'a'] == 0)
                            tot--;
                            
                            i++;
                                 }
                            
                            j++;
                          }
                         
                   }
                   if(ans[k-1] == 0)
                   break;
                   
                   else
                   {
                          ass++;
                          str.append(ans[k-1]);
                          str.append(System.lineSeparator());
                   }
              }
              System.out.println(ass);
              System.out.println(str);
       }
}