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 final class string_div
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	
    public static void main(String args[]) throws Exception
    {
		char[] a=sc.next().toCharArray();int n=a.length;int[][] pre=new int[26][n];int[] sum=new int[26];
		
		for(int i=0;i<n;i++)
		{
			sum[a[i]-'a']++;
			
			for(int j=0;j<26;j++)
			{
				pre[j][i]=sum[j];
			}
		}
		
		long[] res=new long[27];
		
		for(int i=1;i<=26;i++)
		{
			int[] cnt=new int[26];int curr=0;
			
			for(int j=0,k=0;j<n;j++)
			{
				while(k<n)
				{
					if(curr==i && cnt[a[k]-'a']==0)
					{
						break;
					}
					
					cnt[a[k]-'a']++;
					
					if(cnt[a[k]-'a']==1)
					{
						curr++;
					}
					k++;
				}
				
				//out.println(i+" "+j+" "+k);
				
				res[i]+=(k-j);cnt[a[j]-'a']--;
				
				if(cnt[a[j]-'a']==0)
				{
					curr--;
				}
			}
		}
		
		int max=0;
		
		for(int i=0;i<26;i++)
		{
			if(sum[i]>0)
			{
				max++;
			}
		}
		
		long curr=0;out.println(max);
		
		for(int i=1;i<=max;i++)
		{
			res[i]-=curr;
			
			out.println(res[i]);
			
			curr+=res[i];
		}
		
		out.close();
    }
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}