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;
const int MAXS = 300011, ALPHA = 31;
char str[MAXS];
int last[ALPHA], l;
long long ans[ALPHA];
struct forsort {
  int a, la;
  friend bool operator<(const forsort& x, const forsort& y) {
    return y.la < x.la;
  }
} fs[ALPHA];
int main() {
  while (scanf("%s", str) > 0) {
    memset(last, 0, sizeof(last));
    memset(ans, 0, sizeof(ans));
    l = strlen(str);
    for (int i = 1; i <= l; i++) {
      for (int a = 0; a < 26; a++) {
        if (a != str[i - 1] - 'a') {
          fs[a + 1].a = a;
          fs[a + 1].la = last[a];
        } else {
          fs[a + 1].la = 0;
        }
      }
      sort(fs + 1, fs + 27);
      fs[0].la = i;
      for (int j = 1; j <= 26; j++) {
        ans[j] += fs[j - 1].la - fs[j].la;
      }
      last[str[i - 1] - 'a'] = i;
    }
    int a = 0;
    for (int i = 0; i < 26; i++)
      if (last[i] > 0) a++;
    printf("%d\n", a);
    for (int i = 1; i <= a; i++) cout << ans[i] << "\n";
  }
  return 0;
}