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;
long long ara[45], tra[35];
int main() {
  long long n, m, i, j, k, sz, fr, sc, p, cnt = 0;
  vector<long long> v;
  string str;
  cin >> str;
  sz = str.size();
  for (i = 1; i <= 26; i++) {
    memset(tra, 0, sizeof tra);
    fr = 0;
    sc = 0;
    cnt = 0;
    while (1) {
      p = str[sc] - 97;
      tra[p]++;
      if (tra[p] == 1) {
        cnt++;
      }
      if (cnt <= i) {
        sc++;
        if (sc == sz) break;
      } else {
        for (j = fr; j < sc; j++) {
          ara[i] = ara[i] + (sc - j);
          p = str[j] - 97;
          tra[p]--;
          if (tra[p] == 0) cnt--;
          if (cnt <= i) break;
        }
        fr = j + 1;
        if (fr == sz) break;
        sc++;
        if (sc == sz) break;
      }
    }
    if (fr < sz) {
      for (j = fr; j < sz; j++) {
        ara[i] = ara[i] + (sz - j);
      }
    }
    if (ara[i] - ara[i - 1] == 0)
      break;
    else {
      v.push_back(ara[i] - ara[i - 1]);
    }
  }
  m = v.size();
  printf("%lld\n", m);
  for (i = 0; i < m; i++) {
    printf("%lld\n", v[i]);
  }
  return 0;
}