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;
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-6;
const int INF = 0x3fffffff;
const long long LINF = INF * 1ll * INF;
const double PI = acos(-1.0);
using namespace std;
char s[300005];
int last[27];
int tmp[55];
long long ans[55];
int main(void) {
  scanf("%s", s);
  set<int> st;
  memset(last, -1, sizeof(last));
  st.insert(-1);
  int mx = 0;
  for (int i = 0; s[i]; i++) {
    if (~last[s[i] - 'a']) st.erase(last[s[i] - 'a']);
    last[s[i] - 'a'] = i;
    int m = 0;
    for (__typeof(st.begin()) it = st.begin(); it != st.end(); it++)
      tmp[m++] = *it;
    int x = i;
    for (int j = m - 1, cnt = 1; j >= 0; j--, cnt++) {
      ans[cnt] += x - tmp[j];
      mx = max(mx, cnt);
      x = tmp[j];
    }
    st.insert(i);
  }
  printf("%d\n", mx);
  for (int i = 1; i <= mx; i++) printf("%I64d\n", ans[i]);
}