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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char a[345678];
int n, c1[26], c2[26], d1, d2, td;
int main() {
  scanf("%s", a);
  n = strlen(a);
  for (int i = 0; i < n; i++) c1[a[i] - 'a']++;
  for (int i = 0; i < 26; i++) td += (c1[i] > 0);
  printf("%d\n", td);
  for (int k = 1; k <= td; k++) {
    long long ans = 0;
    d1 = d2 = 0;
    for (int i = 0; i < 26; i++) c1[i] = c2[i] = 0;
    for (int i = 0, s = 0, e = -1; i < n; i++) {
      while (s < n && d1 < k) {
        d1 += !c1[a[s] - 'a'];
        c1[a[s] - 'a']++;
        s++;
      }
      while (e + 1 < n && d2 + (!c2[a[e + 1] - 'a']) <= k) {
        d2 += !c2[a[e + 1] - 'a'];
        c2[a[e + 1] - 'a']++;
        e++;
      }
      if (d1 == k && d2 == k) ans += e - s + 2;
      if (c1[a[i] - 'a']-- == 1) d1--;
      if (c2[a[i] - 'a']-- == 1) d2--;
    }
    printf("%lld\n", ans);
  }
  return 0;
}