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;
int n;
int cnt[30];
char s[300005];
int add(int x, int val) {
  int ans = 0;
  if (cnt[x] == 0 && val == 1) ans = 1;
  if (cnt[x] == 1 && val == -1) ans = -1;
  cnt[x] += val;
  return ans;
}
int main() {
  int i, mx, x, l, now;
  long long ans, last;
  scanf(" %s", s);
  n = strlen(s);
  for (i = 0; i < n; i++) cnt[s[i] - 'a'] = 1;
  mx = 0;
  for (i = 0; i < 26; i++) mx += cnt[i];
  printf("%d\n", mx);
  last = 0;
  for (x = 1; x <= mx; x++) {
    ans = 0;
    memset(cnt, 0, sizeof(cnt));
    l = 0;
    now = 0;
    for (i = 0; i < n; i++) {
      now += add(s[i] - 'a', 1);
      while (now > x) {
        now += add(s[l] - 'a', -1);
        l++;
      }
      ans += (long long)i - l + 1;
    }
    printf("%I64d\n", ans - last);
    last = ans;
  }
}