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 = 300000 + 10;
char s[MAX_N];
int cc[MAX_N][26];
inline int calc(int l, int r) {
  int ans = 0;
  for (int i = 0; i < 26; i++)
    if (cc[r][i] - cc[l - 1][i] > 0) ans++;
  return ans;
}
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  unordered_set<char> st;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 26; j++) cc[i][j] = cc[i - 1][j];
    st.insert(s[i]);
    cc[i][s[i] - 'a']++;
  }
  int cnt = (int)st.size();
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    st.clear();
    int lb = 1, rb = 1;
    long long ans = 0LL;
    for (int j = 1; j <= n; j++) {
      st.insert(s[j]);
      if ((int)st.size() >= i) {
        while (lb < j && calc(lb, j) > i) lb++;
        while (rb < j && calc(rb + 1, j) >= i) rb++;
        ans += rb - lb + 1;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}