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;
char S[300010];
long long N, i, DP[30], C[30], j, k, diff, SOL, afis[30];
int main() {
  cin >> (S + 1);
  N = strlen((S + 1));
  for (k = 1; k <= 26; k++) {
    diff = 0;
    for (i = 1; i <= 26; i++) C[i] = 0;
    i = j = 1;
    while (j <= N) {
      if (!C[S[j] - 'a' + 1]) diff++;
      C[S[j] - 'a' + 1]++;
      while (diff > k) {
        C[S[i] - 'a' + 1]--;
        if (!C[S[i] - 'a' + 1]) diff--;
        i = i + 1;
      }
      if (diff <= k) {
        DP[k] = DP[k] + j - i + 1;
      }
      j++;
    }
  }
  SOL = 26;
  for (i = 1; i <= 26; i++) {
    afis[i] = DP[i] - DP[i - 1];
    if (!afis[i]) {
      SOL = i - 1;
      break;
    }
  }
  cout << SOL << "\n";
  for (i = 1; i <= SOL; i++) cout << afis[i] << "\n";
  return 0;
}