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;
int A[255], StrDiv;
long long Am[30];
string S;
int main() {
  cin >> S;
  for (int div = 1; div <= 'z' - 'a' + 1; div++) {
    int a = 0, b = 0, d = 0;
    long long mult = 1;
    for (int i = 'a'; i <= 'z'; i++) A[i] = 0;
    while (b < S.length()) {
      A[S[b]]++;
      if (A[S[b]] == 1) d++;
      b++;
      while (d > div) {
        mult = 1;
        A[S[a]]--;
        if (A[S[a]] == 0) d--;
        a++;
      }
      while (d == div && A[S[a]] - 1 > 0) {
        A[S[a]]--;
        a++;
        mult++;
      }
      if (d == div) Am[div] += mult;
    }
    if (Am[div]) StrDiv = div;
  }
  cout << StrDiv << "\n";
  for (int d = 1; d <= StrDiv; d++) cout << Am[d] << "\n";
}