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 double ep = 1e-12;
char str[333333];
int main() {
  set<long long> seet;
  map<char, long long> maap;
  while (scanf("%s", &str) != EOF) {
    long long len = strlen(str);
    seet.clear();
    for (long long i = 0; i < len; i++) seet.insert(str[i]);
    long long dive = seet.size();
    printf("%lld\n", dive);
    map<char, long long> m1;
    map<char, long long> m2;
    for (long long i = 1; i <= dive; i++) {
      long long ans = 0, p = 1, q = 1;
      m1.clear();
      m2.clear();
      m1[str[0]]++;
      m2[str[0]]++;
      long long d1 = 1, d2 = 1;
      for (long long l = 1; l <= len; ++l) {
        if (l > 1) {
          m1[str[l - 2]]--;
          if (m1[str[l - 2]] == 0) d1--;
          m2[str[l - 2]]--;
          if (m2[str[l - 2]] == 0) d2--;
        }
        while (d1 < i && p <= len - 1) {
          if (m1[str[p]] == 0) d1++;
          m1[str[p++]]++;
        }
        while (d2 <= i && q <= len) {
          if (m2[str[q]] == 0) d2++;
          m2[str[q++]]++;
        }
        ans += q - p - (d1 < i);
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}