#include <bits/stdc++.h>
using namespace std;
char s[300001];
int k, c[128], m;
int main() {
  scanf("%s", s);
  k = set<char>(s, s + 300001).size() - 1;
  printf("%d\n", k);
  for (int i = 1; i <= k; ++i) {
    long long b = 0;
    int j = 0, l = 0;
    for (int z = 0; s[z]; ++z) {
      while (m < i && s[j]) m += !c[s[j++]]++;
      while (c[s[l]]) ++l;
      if (m == i) b += l - j + 1;
      m -= !--c[s[z]];
    }
    printf("%I64d\n", b);
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
char s[300001];
int k, c[128], m;
int main() {
  scanf("%s", s);
  k = set<char>(s, s + 300001).size() - 1;
  printf("%d\n", k);
  for (int i = 1; i <= k; ++i) {
    long long b = 0;
    int j = 0, l = 0;
    for (int z = 0; s[z]; ++z) {
      while (m < i && s[j]) m += !c[s[j++]]++;
      while (c[s[l]]) ++l;
      if (m == i) b += l - j + 1;
      m -= !--c[s[z]];
    }
    printf("%I64d\n", b);
  }
}

```