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 N = 3 * 100000 + 10000;
char s[N];
int last[30];
long long ans[30];
long long p[30];
int main() {
  scanf("%s", s);
  int len = strlen(s);
  memset(last, -1, sizeof(last));
  memset(ans, 0, sizeof(ans));
  int cnt = 0;
  for (int i = 0; i < len; i++) {
    last[s[i] - 'a'] = i;
    cnt = 0;
    for (int j = 0; j < 26; j++)
      if (last[j] != -1) {
        p[cnt++] = last[j];
      }
    sort(p, p + cnt);
    for (int j = 1; j < cnt; j++) ans[cnt - j] += p[j] - p[j - 1];
    ans[cnt] += p[0] + 1;
  }
  for (int i = 26; i >= 1; i--)
    if (ans[i] > 0) {
      printf("%d\n", i);
      for (int j = 1; j <= i; j++) printf("%I64d\n", ans[j]);
      break;
    }
  return 0;
}