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 ans[300005];
int add[300005], pos[300005];
char str[300005];
int main() {
  cin >> str;
  memset(pos, -1, sizeof(pos));
  int len = strlen(str);
  int i;
  for (i = 0; i < len; ++i) {
    if (pos[str[i] - 'a' + 1] == -1) {
      for (int j = 26; j >= 1; --j) add[j] = add[j - 1];
      add[1]++;
    } else {
      int k = i - pos[str[i] - 'a' + 1] - 1;
      int flag;
      for (flag = 1; k >= 1; flag++) {
        k -= add[flag];
      }
      add[flag] += add[flag - 1];
      for (int j = flag - 1; j >= 1; j--) add[j] = add[j - 1];
      add[1]++;
    }
    pos[str[i] - 'a' + 1] = i;
    for (int j = 26; j >= 1; j--) {
      ans[j] += add[j];
    }
  }
  int j;
  for (j = 26; j >= 1; j--) {
    if (ans[j]) {
      printf("%d\n", j);
      break;
    }
  }
  for (int i = 1; i <= j; ++i) printf("%lld\n", ans[i]);
  return 0;
}