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 Test {
	public static void main(String[] args) throws IOException {
		BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		String S = rd.readLine();
		int n = S.length();
		positions = new ArrayList[26];
		for (int i = 0; i < 26; i++) {
			positions[i] = new ArrayList<Integer>();
		}
		for (int i = 0; i < n; i++) {
			int ch = (int) (S.charAt(i) - 'a');
			positions[ch].add(i);
		}

		long[] freq = new long[27];
		currentPos = new int[26];
		for (int i = 0; i < n; i++) {
			ArrayList<Integer> letterPositions = new ArrayList<Integer>();
			for (int ch = 0; ch < 26; ch++) {
				int minimalIndex = findMinimalIndexOf(ch, i);
				if (minimalIndex == -1)
					continue;
				letterPositions.add(minimalIndex);
			}
			Collections.sort(letterPositions);

			int k = letterPositions.size();
			for (int x = 0; x < k; x++) {
				freq[x + 1] += sizeOfInterval(letterPositions.get(x),
						(x < k - 1 ? letterPositions.get(x + 1) - 1 : n - 1));
			}
		}
		int diver = 0;
		for (int i = 26; i >= 0; i--) {
			if (freq[i] > 0) {
				diver = i;
				break;
			}
		}
		pw.println(diver);
		for (int i = 1; i <= diver; i++) {
			pw.println(freq[i]);
		}
		pw.flush();
		pw.close();
	}

	static int sizeOfInterval(int a, int b) {
		if (b < a)
			return 0;
		return b - a + 1;
	}

	static int findMinimalIndexOf(int ch, int afterIndex) {
		if (currentPos[ch] >= positions[ch].size())
			return -1; // not found
		for (int i = currentPos[ch]; i < positions[ch].size(); i++) {
			int position = positions[ch].get(i);
			if (position >= afterIndex) {
				currentPos[ch] = i;
				return position;
			}
		}
		currentPos[ch] = positions[ch].size();
		return -1; // not found
	}

	static ArrayList<Integer>[] positions;
	static int[] currentPos;

}