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 int MAX_N = 320005;
set<int> st;
char ch[MAX_N];
int pos[30];
long long ans[30];
int main() {
  scanf("%s", ch + 1);
  int len = strlen(ch + 1);
  for (int i = 1; i <= len; i++) {
    int c = ch[i] - 'a';
    if (pos[c]) st.erase(pos[c]);
    st.insert(i);
    pos[c] = i;
    int pre = 0, d = st.size();
    for (set<int>::iterator it = st.begin(); it != st.end(); ++it) {
      ans[d] += 1ll * (*it - pre);
      pre = *it;
      d--;
    }
  }
  int diversity;
  for (int i = 1; i < 30; i++)
    if (ans[i] == 0) {
      diversity = i - 1;
      break;
    }
  printf("%d\n", diversity);
  for (int i = 1; i <= diversity; i++) cout << ans[i] << endl;
  return 0;
}