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>
const int N = 300240, inf = 0x7fffffff;
using namespace std;
char s[N];
long long cnt[N][28], last[26], ans[26];
bool used[26];
int main(void) {
  int n, tot, p;
  while (scanf("%s", s + 1) != EOF) {
    n = strlen(s + 1);
    tot = 0;
    memset(used, 0, sizeof(used));
    for (int i = 1; i <= n; i++) used[s[i] - 'a'] = true;
    for (int i = 0; i < 26; i++)
      if (used[i]) tot++;
    memset(last, 0, sizeof(last));
    memset(ans, 0, sizeof(ans));
    for (int i = n; i; i--) {
      last[s[i] - 'a'] = i;
      for (int j = 0; j < 26; j++) {
        cnt[i][j] = last[j];
      }
    }
    for (int i = 1; i <= n; i++) {
      sort(cnt[i], cnt[i] + 26);
      for (int j = 0; j < 26; j++)
        if (cnt[i][j] == i) {
          p = j;
          break;
        }
      for (int j = p + 1; j < 26; j++) {
        ans[j - p] += cnt[i][j] - cnt[i][j - 1];
      }
      ans[26 - p] += n - cnt[i][25] + 1;
    }
    printf("%d\n", tot);
    for (int i = 1; i <= tot; i++) printf("%I64d\n", ans[i]);
  }
  return 0;
}